From f8004f9b41bc011f8e961f3065dbab61f1f63a7d Mon Sep 17 00:00:00 2001
From: simonhancrew <597494370@qq.com>
Date: Thu, 27 May 2021 15:58:22 +0800
Subject: [PATCH 01/95] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200977.=E6=9C=89?=
=?UTF-8?q?=E5=BA=8F=E6=95=B0=E7=BB=84=E7=9A=84=E5=B9=B3=E6=96=B9=20Python?=
=?UTF-8?q?3=20Go=20Rust=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0977.有序数组的平方.md | 60 +++++++++++++++++++++++++-
1 file changed, 58 insertions(+), 2 deletions(-)
diff --git a/problems/0977.有序数组的平方.md b/problems/0977.有序数组的平方.md
index 167a258c..b5d392e6 100644
--- a/problems/0977.有序数组的平方.md
+++ b/problems/0977.有序数组的平方.md
@@ -100,10 +100,66 @@ public:
Java:
Python:
-
+```Python
+class Solution:
+ def sortedSquares(self, nums: List[int]) -> List[int]:
+ n = len(nums)
+ i,j,k = 0,n - 1,n - 1
+ ans = [-1] * n
+ while i <= j:
+ lm = nums[i] ** 2
+ rm = nums[j] ** 2
+ if lm > rm:
+ ans[k] = lm
+ i += 1
+ else:
+ ans[k] = rm
+ j -= 1
+ k -= 1
+ return ans
+```
Go:
-
+```Go
+func sortedSquares(nums []int) []int {
+ n := len(nums)
+ i, j, k := 0, n-1, n-1
+ ans := make([]int, n)
+ for i <= j {
+ lm, rm := nums[i]*nums[i], nums[j]*nums[j]
+ if lm > rm {
+ ans[k] = lm
+ i++
+ } else {
+ ans[k] = rm
+ j--
+ }
+ k--
+ }
+ return ans
+}
+```
+Rust
+```
+impl Solution {
+ pub fn sorted_squares(nums: Vec) -> Vec {
+ let n = nums.len();
+ let (mut i,mut j,mut k) = (0,n - 1,n- 1);
+ let mut ans = vec![0;n];
+ while i <= j{
+ if nums[i] * nums[i] < nums[j] * nums[j] {
+ ans[k] = nums[j] * nums[j];
+ j -= 1;
+ }else{
+ ans[k] = nums[i] * nums[i];
+ i += 1;
+ }
+ k -= 1;
+ }
+ ans
+ }
+}
+```
-----------------------
* 作者微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
From 597f24d2a07341d2dadcc1ecbfdbff372863377c Mon Sep 17 00:00:00 2001
From: boom-jumper <56831966+boom-jumper@users.noreply.github.com>
Date: Thu, 27 May 2021 20:30:37 +0800
Subject: [PATCH 02/95] =?UTF-8?q?=E6=B7=BB=E5=8A=A00232=E7=94=A8=E6=A0=88?=
=?UTF-8?q?=E5=AE=9E=E7=8E=B0=E9=98=9F=E5=88=97=20python3=20=E7=89=88?=
=?UTF-8?q?=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0232.用栈实现队列.md | 44 +++++++++++++++++++++++++++++
1 file changed, 44 insertions(+)
diff --git a/problems/0232.用栈实现队列.md b/problems/0232.用栈实现队列.md
index 10314a08..c2af71f7 100644
--- a/problems/0232.用栈实现队列.md
+++ b/problems/0232.用栈实现队列.md
@@ -282,6 +282,50 @@ class MyQueue {
Python:
+```python
+# 使用两个栈实现先进先出的队列
+class MyQueue:
+ def __init__(self):
+ """
+ Initialize your data structure here.
+ """
+ self.stack1 = list()
+ self.stack2 = list()
+
+ def push(self, x: int) -> None:
+ """
+ Push element x to the back of queue.
+ """
+ # self.stack1用于接受元素
+ self.stack1.append(x)
+
+ def pop(self) -> int:
+ """
+ Removes the element from in front of queue and returns that element.
+ """
+ # self.stack2用于弹出元素,如果self.stack2为[],则将self.stack1中元素全部弹出给self.stack2
+ if self.stack2 == []:
+ while self.stack1:
+ tmp = self.stack1.pop()
+ self.stack2.append(tmp)
+ return self.stack2.pop()
+
+ def peek(self) -> int:
+ """
+ Get the front element.
+ """
+ if self.stack2 == []:
+ while self.stack1:
+ tmp = self.stack1.pop()
+ self.stack2.append(tmp)
+ return self.stack2[-1]
+
+ def empty(self) -> bool:
+ """
+ Returns whether the queue is empty.
+ """
+ return self.stack1 == [] and self.stack2 == []
+```
Go:
From 0ae9d1808cf4db850f28ad9cfa837506bfec9588 Mon Sep 17 00:00:00 2001
From: boom-jumper <56831966+boom-jumper@users.noreply.github.com>
Date: Thu, 27 May 2021 20:37:53 +0800
Subject: [PATCH 03/95] =?UTF-8?q?=E6=B7=BB=E5=8A=A00150=E9=80=86=E6=B3=A2?=
=?UTF-8?q?=E5=85=B0=E8=A1=A8=E8=BE=BE=E5=BC=8F=E6=B1=82=E5=80=BC=20python?=
=?UTF-8?q?3=20=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0150.逆波兰表达式求值.md | 16 ++++++++++++++++
1 file changed, 16 insertions(+)
diff --git a/problems/0150.逆波兰表达式求值.md b/problems/0150.逆波兰表达式求值.md
index e38bc1a3..c8b0da08 100644
--- a/problems/0150.逆波兰表达式求值.md
+++ b/problems/0150.逆波兰表达式求值.md
@@ -224,6 +224,22 @@ var evalRPN = function(tokens) {
};
```
+python3
+
+```python
+def evalRPN(tokens) -> int:
+ stack = list()
+ for i in range(len(tokens)):
+ if tokens[i] not in ["+", "-", "*", "/"]:
+ stack.append(tokens[i])
+ else:
+ tmp1 = stack.pop()
+ tmp2 = stack.pop()
+ res = eval(tmp2+tokens[i]+tmp1)
+ stack.append(str(int(res)))
+ return stack[-1]
+```
+
-----------------------
* 作者微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
From d7235d9f51b532dda387bf6e00be225f931ed275 Mon Sep 17 00:00:00 2001
From: NevS <1173325467@qq.com>
Date: Thu, 27 May 2021 21:39:32 +0800
Subject: [PATCH 04/95] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=2002.07.=E9=93=BE?=
=?UTF-8?q?=E8=A1=A8=E7=9B=B8=E4=BA=A4=20go=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/面试题02.07.链表相交.md | 36 +++++++++++++++++++++++++
1 file changed, 36 insertions(+)
diff --git a/problems/面试题02.07.链表相交.md b/problems/面试题02.07.链表相交.md
index 9e97264c..13014dd1 100644
--- a/problems/面试题02.07.链表相交.md
+++ b/problems/面试题02.07.链表相交.md
@@ -155,6 +155,42 @@ Python:
Go:
+```go
+func getIntersectionNode(headA, headB *ListNode) *ListNode {
+ curA := headA
+ curB := headB
+ lenA, lenB := 0, 0
+ // 求A,B的长度
+ for curA != nil {
+ curA = curA.Next
+ lenA++
+ }
+ for curB != nil {
+ curB = curB.Next
+ lenB++
+ }
+ var step int
+ var fast, slow *ListNode
+ // 请求长度差,并且让更长的链表先走相差的长度
+ if lenA > lenB {
+ step = lenA - lenB
+ fast, slow = headA, headB
+ } else {
+ step = lenB - lenA
+ fast, slow = headB, headA
+ }
+ for i:=0; i < step; i++ {
+ fast = fast.Next
+ }
+ // 遍历两个链表遇到相同则跳出遍历
+ for fast != slow {
+ fast = fast.Next
+ slow = slow.Next
+ }
+ return fast
+}
+```
+
javaScript:
```js
From 7f3b661add171dab5cee5ebae5516962cf55cf3e Mon Sep 17 00:00:00 2001
From: zqh1059405318 <1059405318@qq.com>
Date: Thu, 27 May 2021 23:44:00 +0800
Subject: [PATCH 05/95] =?UTF-8?q?=E6=9B=B4=E6=96=B00541.=E5=8F=8D=E8=BD=AC?=
=?UTF-8?q?=E5=AD=97=E7=AC=A6=E4=B8=B2II=20Java=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0541.反转字符串II.md | 33 ++++++++++++++----------------
1 file changed, 15 insertions(+), 18 deletions(-)
diff --git a/problems/0541.反转字符串II.md b/problems/0541.反转字符串II.md
index db53caf7..6c8f3e94 100644
--- a/problems/0541.反转字符串II.md
+++ b/problems/0541.反转字符串II.md
@@ -106,27 +106,24 @@ Java:
class Solution {
public String reverseStr(String s, int k) {
StringBuffer res = new StringBuffer();
-
- for (int i = 0; i < s.length(); i += (2 * k)) {
+ int length = s.length();
+ int start = 0;
+ while (start < length) {
+ // 找到k处和2k处
StringBuffer temp = new StringBuffer();
- // 剩余字符大于 k 个,每隔 2k 个字符的前 k 个字符进行反转
- if (i + k <= s.length()) {
- // 反转前 k 个字符
- temp.append(s.substring(i, i + k));
- res.append(temp.reverse());
+ // 与length进行判断,如果大于length了,那就将其置为length
+ int firstK = (start + k > length) ? length : start + k;
+ int secondK = (start + (2 * k) > length) ? length : start + (2 * k);
- // 反转完前 k 个字符之后,如果紧接着还有 k 个字符,则直接加入这 k 个字符
- if (i + 2 * k <= s.length()) {
- res.append(s.substring(i + k, i + 2 * k));
- // 不足 k 个字符,则直接加入剩下所有字符
- } else {
- res.append(s.substring(i + k, s.length()));
- }
- continue;
- }
- // 剩余字符少于 k 个,则将剩余字符全部反转。
- temp.append(s.substring(i, s.length()));
+ //无论start所处位置,至少会反转一次
+ temp.append(s.substring(start, firstK));
res.append(temp.reverse());
+
+ // 如果firstK到secondK之间有元素,这些元素直接放入res里即可。
+ if (firstK < secondK) { //此时剩余长度一定大于k。
+ res.append(s.substring(firstK, secondK));
+ }
+ start += (2 * k);
}
return res.toString();
}
From ce1a80b0166b8f37c03c3ed867dc6d89279f345a Mon Sep 17 00:00:00 2001
From: jojoo15 <75017412+jojoo15@users.noreply.github.com>
Date: Thu, 27 May 2021 18:01:48 +0200
Subject: [PATCH 06/95] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200078.=E5=AD=90?=
=?UTF-8?q?=E9=9B=86=20python3=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
添加 0078.子集 python3版本
---
problems/0078.子集.md | 15 ++++++++++++++-
1 file changed, 14 insertions(+), 1 deletion(-)
diff --git a/problems/0078.子集.md b/problems/0078.子集.md
index 17c4fb5c..0b2f3c09 100644
--- a/problems/0078.子集.md
+++ b/problems/0078.子集.md
@@ -205,7 +205,20 @@ class Solution {
```
Python:
-
+```python3
+class Solution:
+ def subsets(self, nums: List[int]) -> List[List[int]]:
+ res = []
+ path = []
+ def backtrack(nums,startIndex):
+ res.append(path[:]) #收集子集,要放在终止添加的上面,否则会漏掉自己
+ for i in range(startIndex,len(nums)): #当startIndex已经大于数组的长度了,就终止了,for循环本来也结束了,所以不需要终止条件
+ path.append(nums[i])
+ backtrack(nums,i+1) #递归
+ path.pop() #回溯
+ backtrack(nums,0)
+ return res
+```
Go:
```Go
From c722642392ab8853f91ffac654149531b8117973 Mon Sep 17 00:00:00 2001
From: NevS <1173325467@qq.com>
Date: Fri, 28 May 2021 00:33:17 +0800
Subject: [PATCH 07/95] =?UTF-8?q?=E6=9B=B4=E6=96=B0=200142.=E7=8E=AF?=
=?UTF-8?q?=E5=BD=A2=E9=93=BE=E8=A1=A8II=20go=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
1、原先的Go版本代码显示错乱,一部分代码未显示
2、去掉多余判断,让代码更简洁
---
problems/0142.环形链表II.md | 28 +++++++++++-----------------
1 file changed, 11 insertions(+), 17 deletions(-)
diff --git a/problems/0142.环形链表II.md b/problems/0142.环形链表II.md
index 15c2d1f9..9deb1e0c 100644
--- a/problems/0142.环形链表II.md
+++ b/problems/0142.环形链表II.md
@@ -234,25 +234,19 @@ class Solution:
```
Go:
-```func detectCycle(head *ListNode) *ListNode {
- if head ==nil{
- return head
- }
- slow:=head
- fast:=head.Next
-
- for fast!=nil&&fast.Next!=nil{
- if fast==slow{
- slow=head
- fast=fast.Next
- for fast!=slow {
- fast=fast.Next
- slow=slow.Next
+```go
+func detectCycle(head *ListNode) *ListNode {
+ slow, fast := head, head
+ for fast != nil && fast.Next != nil {
+ slow = slow.Next
+ fast = fast.Next.Next
+ if slow == fast {
+ for slow != head {
+ slow = slow.Next
+ head = head.Next
}
- return slow
+ return head
}
- fast=fast.Next.Next
- slow=slow.Next
}
return nil
}
From b41866f1f4a2fcc7c631bbaf9de5166465a7586c Mon Sep 17 00:00:00 2001
From: jojoo15 <75017412+jojoo15@users.noreply.github.com>
Date: Thu, 27 May 2021 18:57:04 +0200
Subject: [PATCH 08/95] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200090.=E5=AD=90?=
=?UTF-8?q?=E9=9B=86=20python3=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
添加 0090.子集 python3版本
---
problems/0090.子集II.md | 18 +++++++++++++++++-
1 file changed, 17 insertions(+), 1 deletion(-)
diff --git a/problems/0090.子集II.md b/problems/0090.子集II.md
index 6b12a95b..71aef5c7 100644
--- a/problems/0090.子集II.md
+++ b/problems/0090.子集II.md
@@ -208,7 +208,23 @@ class Solution {
```
Python:
-
+```python3
+class Solution:
+ def subsetsWithDup(self, nums: List[int]) -> List[List[int]]:
+ res = [] #存放符合条件结果的集合
+ path = [] #用来存放符合条件结果
+ def backtrack(nums,startIndex):
+ res.append(path[:])
+ for i in range(startIndex,len(nums)):
+ if i > startIndex and nums[i] == nums[i - 1]: #我们要对同一树层使用过的元素进行跳过
+ continue
+ path.append(nums[i])
+ backtrack(nums,i+1) #递归
+ path.pop() #回溯
+ nums = sorted(nums) #去重需要排序
+ backtrack(nums,0)
+ return res
+```
Go:
```Go
From 60624686a894dadcfefb79cd38e40f0308217f93 Mon Sep 17 00:00:00 2001
From: jojoo15 <75017412+jojoo15@users.noreply.github.com>
Date: Fri, 28 May 2021 00:09:28 +0200
Subject: [PATCH 09/95] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200491.=E9=80=92?=
=?UTF-8?q?=E5=A2=9E=E5=AD=90=E5=BA=8F=E5=88=97=20python3=E7=89=88?=
=?UTF-8?q?=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
添加 0491.递增子序列 python3版本
---
problems/0491.递增子序列.md | 23 ++++++++++++++++++++++-
1 file changed, 22 insertions(+), 1 deletion(-)
diff --git a/problems/0491.递增子序列.md b/problems/0491.递增子序列.md
index 691f7aef..5538a2c9 100644
--- a/problems/0491.递增子序列.md
+++ b/problems/0491.递增子序列.md
@@ -229,7 +229,28 @@ class Solution {
Python:
-
+```python3
+class Solution:
+ def findSubsequences(self, nums: List[int]) -> List[List[int]]:
+ res = []
+ path = []
+ def backtrack(nums,startIndex):
+ repeat = [] #这里使用数组来进行去重操作
+ if len(path) >=2:
+ res.append(path[:]) #注意这里不要加return,要取树上的节点
+ for i in range(startIndex,len(nums)):
+ if nums[i] in repeat:
+ continue
+ if len(path) >= 1:
+ if nums[i] < path[-1]:
+ continue
+ repeat.append(nums[i]) #记录这个元素在本层用过了,本层后面不能再用了
+ path.append(nums[i])
+ backtrack(nums,i+1)
+ path.pop()
+ backtrack(nums,0)
+ return res
+```
Go:
From 6c33729d9f15ea6db9f9497c70bb87081394b2c1 Mon Sep 17 00:00:00 2001
From: fusunx <1102654482@qq.com>
Date: Fri, 28 May 2021 08:11:56 +0800
Subject: [PATCH 10/95] =?UTF-8?q?0055.=E8=B7=B3=E8=B7=83=E6=B8=B8=E6=88=8F?=
=?UTF-8?q?.md=20Javascript?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0055.跳跃游戏.md | 15 ++++++++++++++-
1 file changed, 14 insertions(+), 1 deletion(-)
diff --git a/problems/0055.跳跃游戏.md b/problems/0055.跳跃游戏.md
index b4b42a4a..8618515e 100644
--- a/problems/0055.跳跃游戏.md
+++ b/problems/0055.跳跃游戏.md
@@ -141,7 +141,20 @@ func canJUmp(nums []int) bool {
}
```
-
+Javascript:
+```Javascript
+var canJump = function(nums) {
+ if(nums.length === 1) return true
+ let cover = 0
+ for(let i = 0; i <= cover; i++) {
+ cover = Math.max(cover, i + nums[i])
+ if(cover >= nums.length - 1) {
+ return true
+ }
+ }
+ return false
+};
+```
-----------------------
From c01a968b92f97c98433432bfabc769c22d5f0f1c Mon Sep 17 00:00:00 2001
From: xll <18574553598@163.com>
Date: Fri, 28 May 2021 11:05:37 +0800
Subject: [PATCH 11/95] =?UTF-8?q?0404=E5=B7=A6=E5=8F=B6=E5=AD=90=E4=B9=8B?=
=?UTF-8?q?=E5=92=8CJavaScript=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0404.左叶子之和.md | 46 +++++++++++++++++++++++++++++++-
1 file changed, 45 insertions(+), 1 deletion(-)
diff --git a/problems/0404.左叶子之和.md b/problems/0404.左叶子之和.md
index 30d702b4..e5daa9db 100644
--- a/problems/0404.左叶子之和.md
+++ b/problems/0404.左叶子之和.md
@@ -226,7 +226,51 @@ class Solution:
```
Go:
-
+JavaScript:
+递归版本
+```javascript
+var sumOfLeftLeaves = function(root) {
+ //采用后序遍历 递归遍历
+ // 1. 确定递归函数参数
+ const nodesSum = function(node){
+ // 2. 确定终止条件
+ if(node===null){
+ return 0;
+ }
+ let leftValue = sumOfLeftLeaves(node.left);
+ let rightValue = sumOfLeftLeaves(node.right);
+ // 3. 单层递归逻辑
+ let midValue = 0;
+ if(node.left&&node.left.left===null&&node.left.right===null){
+ midValue = node.left.val;
+ }
+ let sum = midValue + leftValue + rightValue;
+ return sum;
+ }
+ return nodesSum(root);
+};
+```
+迭代版本
+```javascript
+var sumOfLeftLeaves = function(root) {
+ //采用层序遍历
+ if(root===null){
+ return null;
+ }
+ let queue = [];
+ let sum = 0;
+ queue.push(root);
+ while(queue.length){
+ let node = queue.shift();
+ if(node.left!==null&&node.left.left===null&&node.left.right===null){
+ sum+=node.left.val;
+ }
+ node.left&&queue.push(node.left);
+ node.right&&queue.push(node.right);
+ }
+ return sum;
+};
+```
From f6a9d649ea71d8d0685423278b8c3879fb8143f4 Mon Sep 17 00:00:00 2001
From: LiangDazhu <42199191+LiangDazhu@users.noreply.github.com>
Date: Fri, 28 May 2021 16:30:11 +0800
Subject: [PATCH 12/95] =?UTF-8?q?Update=200343.=E6=95=B4=E6=95=B0=E6=8B=86?=
=?UTF-8?q?=E5=88=86.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
Added python version code
有一行文本我觉得写重复了, 给删掉了
---
problems/0343.整数拆分.md | 19 +++++++++++++------
1 file changed, 13 insertions(+), 6 deletions(-)
diff --git a/problems/0343.整数拆分.md b/problems/0343.整数拆分.md
index c9423a1a..fefaa293 100644
--- a/problems/0343.整数拆分.md
+++ b/problems/0343.整数拆分.md
@@ -51,10 +51,6 @@ dp[i]的定义讲贯彻整个解题过程,下面哪一步想不懂了,就想
**那有同学问了,j怎么就不拆分呢?**
-j是从1开始遍历,拆分j的情况,在遍历j的过程中其实都计算过了。
-
-**那有同学问了,j怎么就不拆分呢?**
-
j是从1开始遍历,拆分j的情况,在遍历j的过程中其实都计算过了。那么从1遍历j,比较(i - j) * j和dp[i - j] * j 取最大的。递推公式:dp[i] = max(dp[i], max((i - j) * j, dp[i - j] * j));
也可以这么理解,j * (i - j) 是单纯的把整数拆分为两个数相乘,而j * dp[i - j]是拆分成两个以及两个以上的个数相乘。
@@ -213,8 +209,19 @@ class Solution {
```
Python:
-
-
+```python
+class Solution:
+ def integerBreak(self, n: int) -> int:
+ dp = [0] * (n + 1)
+ dp[2] = 1
+ for i in range(3, n + 1):
+ # 假设对正整数 i 拆分出的第一个正整数是 j(1 <= j < i),则有以下两种方案:
+ # 1) 将 i 拆分成 j 和 i−j 的和,且 i−j 不再拆分成多个正整数,此时的乘积是 j * (i-j)
+ # 2) 将 i 拆分成 j 和 i−j 的和,且 i−j 继续拆分成多个正整数,此时的乘积是 j * dp[i-j]
+ for j in range(1, i):
+ dp[i] = max(dp[i], max(j * (i - j), j * dp[i - j]))
+ return dp[n]
+```
Go:
From f9686c7d84422f9403ded97760c8b2e2d653e090 Mon Sep 17 00:00:00 2001
From: jojoo15 <75017412+jojoo15@users.noreply.github.com>
Date: Fri, 28 May 2021 11:19:43 +0200
Subject: [PATCH 13/95] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200046.=E5=85=A8?=
=?UTF-8?q?=E6=8E=92=E5=88=97=20python3=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
添加 0046.全排列 python3版本
---
problems/0046.全排列.md | 21 ++++++++++++++++++++-
1 file changed, 20 insertions(+), 1 deletion(-)
diff --git a/problems/0046.全排列.md b/problems/0046.全排列.md
index 0beb45cf..6e5b528e 100644
--- a/problems/0046.全排列.md
+++ b/problems/0046.全排列.md
@@ -182,7 +182,26 @@ class Solution {
```
Python:
-
+```python3
+class Solution:
+ def permute(self, nums: List[int]) -> List[List[int]]:
+ res = [] #存放符合条件结果的集合
+ path = [] #用来存放符合条件的结果
+ used = [] #用来存放已经用过的数字
+ def backtrack(nums,used):
+ if len(path) == len(nums):
+ return res.append(path[:]) #此时说明找到了一组
+ for i in range(0,len(nums)):
+ if nums[i] in used:
+ continue #used里已经收录的元素,直接跳过
+ path.append(nums[i])
+ used.append(nums[i])
+ backtrack(nums,used)
+ used.pop()
+ path.pop()
+ backtrack(nums,used)
+ return res
+```
Go:
```Go
From 14221a52fe02d489330b6fc254d187de99c99ae2 Mon Sep 17 00:00:00 2001
From: jojoo15 <75017412+jojoo15@users.noreply.github.com>
Date: Fri, 28 May 2021 11:26:43 +0200
Subject: [PATCH 14/95] =?UTF-8?q?=E6=9B=B4=E6=96=B0=200046.=E5=85=A8?=
=?UTF-8?q?=E6=8E=92=E5=88=97=20python3=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
更新 0046.全排列 python3版本,比之前那个更简洁一点,少了个used数组
---
problems/0046.全排列.md | 18 +++++++++++++++++-
1 file changed, 17 insertions(+), 1 deletion(-)
diff --git a/problems/0046.全排列.md b/problems/0046.全排列.md
index 0beb45cf..3ead73e3 100644
--- a/problems/0046.全排列.md
+++ b/problems/0046.全排列.md
@@ -182,7 +182,23 @@ class Solution {
```
Python:
-
+```python3
+class Solution:
+ def permute(self, nums: List[int]) -> List[List[int]]:
+ res = [] #存放符合条件结果的集合
+ path = [] #用来存放符合条件的结果
+ def backtrack(nums):
+ if len(path) == len(nums):
+ return res.append(path[:]) #此时说明找到了一组
+ for i in range(0,len(nums)):
+ if nums[i] in path: #path里已经收录的元素,直接跳过
+ continue
+ path.append(nums[i])
+ backtrack(nums) #递归
+ path.pop() #回溯
+ backtrack(nums)
+ return res
+```
Go:
```Go
From 60d89d920053650736e17fa814d020deedcda89e Mon Sep 17 00:00:00 2001
From: NevS <1173325467@qq.com>
Date: Fri, 28 May 2021 22:59:27 +0800
Subject: [PATCH 15/95] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200349.=E4=B8=A4?=
=?UTF-8?q?=E4=B8=AA=E6=95=B0=E7=BB=84=E7=9A=84=E4=BA=A4=E9=9B=86=20go?=
=?UTF-8?q?=E7=89=88?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0349.两个数组的交集.md | 17 +++++++++++++++++
1 file changed, 17 insertions(+)
diff --git a/problems/0349.两个数组的交集.md b/problems/0349.两个数组的交集.md
index fe019a0a..090480a4 100644
--- a/problems/0349.两个数组的交集.md
+++ b/problems/0349.两个数组的交集.md
@@ -132,6 +132,23 @@ class Solution:
Go:
+```go
+func intersection(nums1 []int, nums2 []int) []int {
+ m := make(map[int]int)
+ for _, v := range nums1 {
+ m[v] = 1
+ }
+ var res []int
+ // 利用count>0,实现重复值只拿一次放入返回结果中
+ for _, v := range nums2 {
+ if count, ok := m[v]; ok && count > 0 {
+ res = append(res, v)
+ m[v]--
+ }
+ }
+ return res
+}
+```
javaScript:
From 1590897824de40843710804d6a0397eca0131ee1 Mon Sep 17 00:00:00 2001
From: LiangDazhu <42199191+LiangDazhu@users.noreply.github.com>
Date: Fri, 28 May 2021 23:33:19 +0800
Subject: [PATCH 16/95] =?UTF-8?q?Update=200096.=E4=B8=8D=E5=90=8C=E7=9A=84?=
=?UTF-8?q?=E4=BA=8C=E5=8F=89=E6=90=9C=E7=B4=A2=E6=A0=91.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
Added python version code
---
problems/0096.不同的二叉搜索树.md | 11 ++++++++++-
1 file changed, 10 insertions(+), 1 deletion(-)
diff --git a/problems/0096.不同的二叉搜索树.md b/problems/0096.不同的二叉搜索树.md
index cee0102f..a9631315 100644
--- a/problems/0096.不同的二叉搜索树.md
+++ b/problems/0096.不同的二叉搜索树.md
@@ -186,7 +186,16 @@ class Solution {
```
Python:
-
+```python
+class Solution:
+ def numTrees(self, n: int) -> int:
+ dp = [0] * (n + 1)
+ dp[0], dp[1] = 1, 1
+ for i in range(2, n + 1):
+ for j in range(1, i + 1):
+ dp[i] += dp[j - 1] * dp[i - j]
+ return dp[-1]
+```
Go:
```Go
From c4de753d6db1548205de14efdb4fc9194d7223c4 Mon Sep 17 00:00:00 2001
From: phoenix
Date: Sat, 29 May 2021 08:07:50 +0800
Subject: [PATCH 17/95] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200020.=E6=9C=89?=
=?UTF-8?q?=E6=95=88=E7=9A=84=E6=8B=AC=E5=8F=B7=20Ruby=20=E7=89=88?=
=?UTF-8?q?=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0015.三数之和.md | 17 +++++++++++++++++
1 file changed, 17 insertions(+)
diff --git a/problems/0015.三数之和.md b/problems/0015.三数之和.md
index 811fc316..4f4ec63a 100644
--- a/problems/0015.三数之和.md
+++ b/problems/0015.三数之和.md
@@ -336,6 +336,23 @@ var threeSum = function(nums) {
```
+ruby:
+```ruby
+def is_valid(strs)
+ symbol_map = {')' => '(', '}' => '{', ']' => '['}
+ stack = []
+ strs.size.times {|i|
+ c = strs[i]
+ if symbol_map.has_key?(c)
+ top_e = stack.shift
+ return false if symbol_map[c] != top_e
+ else
+ stack.unshift(c)
+ end
+ }
+ stack.empty?
+end
+```
-----------------------
* 作者微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
From bd95f6a248f9175216582225500e3c074f0c4180 Mon Sep 17 00:00:00 2001
From: phoenix
Date: Sat, 29 May 2021 09:12:26 +0800
Subject: [PATCH 18/95] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200027.=E7=A7=BB?=
=?UTF-8?q?=E9=99=A4=E5=85=83=E7=B4=A0=20Ruby=20=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0027.移除元素.md | 14 ++++++++++++++
1 file changed, 14 insertions(+)
diff --git a/problems/0027.移除元素.md b/problems/0027.移除元素.md
index 8da0fb89..f0f61d06 100644
--- a/problems/0027.移除元素.md
+++ b/problems/0027.移除元素.md
@@ -186,6 +186,20 @@ var removeElement = (nums, val) => {
};
```
+Ruby:
+```ruby
+def remove_element(nums, val)
+ i = 0
+ nums.each_index do |j|
+ if nums[j] != val
+ nums[i] = nums[j]
+ i+=1
+ end
+ end
+ i
+end
+```
+
-----------------------
* 作者微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
* B站视频:[代码随想录](https://space.bilibili.com/525438321)
From 5512ccdb63370be27a387acab0c1ae366607b089 Mon Sep 17 00:00:00 2001
From: xll <18574553598@163.com>
Date: Sat, 29 May 2021 10:44:58 +0800
Subject: [PATCH 19/95] =?UTF-8?q?0513=E4=BA=8C=E5=8F=89=E6=A0=91=E5=B7=A6?=
=?UTF-8?q?=E5=B0=8F=E8=A7=92=E7=9A=84=E5=80=BCJavaScript=E7=89=88?=
=?UTF-8?q?=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0513.找树左下角的值.md | 47 ++++++++++++++++++++++++++
1 file changed, 47 insertions(+)
diff --git a/problems/0513.找树左下角的值.md b/problems/0513.找树左下角的值.md
index 5c9613e5..97464e3d 100644
--- a/problems/0513.找树左下角的值.md
+++ b/problems/0513.找树左下角的值.md
@@ -298,6 +298,53 @@ class Solution:
```
Go:
+JavaScript:
+1. 递归版本
+```javascript
+var findBottomLeftValue = function(root) {
+ //首先考虑递归遍历 前序遍历 找到最大深度的叶子节点即可
+ let maxPath = 0,resNode = null;
+ // 1. 确定递归函数的函数参数
+ const dfsTree = function(node,curPath){
+ // 2. 确定递归函数终止条件
+ if(node.left===null&&node.right===null){
+ if(curPath>maxPath){
+ maxPath = curPath;
+ resNode = node.val;
+ }
+ // return ;
+ }
+ node.left&&dfsTree(node.left,curPath+1);
+ node.right&&dfsTree(node.right,curPath+1);
+ }
+ dfsTree(root,1);
+ return resNode;
+};
+```
+2. 层序遍历
+```javascript
+var findBottomLeftValue = function(root) {
+ //考虑层序遍历 记录最后一行的第一个节点
+ let queue = [];
+ if(root===null){
+ return null;
+ }
+ queue.push(root);
+ let resNode;
+ while(queue.length){
+ let length = queue.length;
+ for(let i=0; i
Date: Sat, 29 May 2021 12:50:49 +0800
Subject: [PATCH 20/95] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=E4=BA=861047-golang?=
=?UTF-8?q?=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
...除字符串中的所有相邻重复项.md | 17 +++++++++++++++++
1 file changed, 17 insertions(+)
diff --git a/problems/1047.删除字符串中的所有相邻重复项.md b/problems/1047.删除字符串中的所有相邻重复项.md
index 9ca08c96..305a287d 100644
--- a/problems/1047.删除字符串中的所有相邻重复项.md
+++ b/problems/1047.删除字符串中的所有相邻重复项.md
@@ -186,6 +186,23 @@ class Solution:
Go:
+```go
+func removeDuplicates(s string) string {
+ var stack []byte
+ for i := 0; i < len(s);i++ {
+ // 栈不空 且 与栈顶元素不等
+ if len(stack) > 0 && stack[len(stack)-1] == s[i] {
+ // 弹出栈顶元素 并 忽略当前元素(s[i])
+ stack = stack[:len(stack)-1]
+ }else{
+ // 入栈
+ stack = append(stack, s[i])
+ }
+ }
+ return string(stack)
+}
+```
+
javaScript:
```js
From fb91b760eb6fa0583210ca260b77f71418787ab8 Mon Sep 17 00:00:00 2001
From: hk27xing <244798299@qq.com>
Date: Sat, 29 May 2021 16:13:32 +0800
Subject: [PATCH 21/95] =?UTF-8?q?=E7=BA=A0=E6=AD=A339.=E7=BB=84=E5=90=88?=
=?UTF-8?q?=E6=80=BB=E5=92=8C=20Java=E7=89=88=E6=9C=AC=E4=BB=A3=E7=A0=81?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0039.组合总和.md | 40 +++++++++++++++--------------------
1 file changed, 17 insertions(+), 23 deletions(-)
diff --git a/problems/0039.组合总和.md b/problems/0039.组合总和.md
index ab118ee0..a83e42f4 100644
--- a/problems/0039.组合总和.md
+++ b/problems/0039.组合总和.md
@@ -237,34 +237,28 @@ public:
Java:
```Java
+// 剪枝优化
class Solution {
- List> lists = new ArrayList<>();
- Deque deque = new LinkedList<>();
-
- public List> combinationSum3(int k, int n) {
- int[] arr = new int[]{1, 2, 3, 4, 5, 6, 7, 8, 9};
- backTracking(arr, n, k, 0);
- return lists;
+ public List> combinationSum(int[] candidates, int target) {
+ List> res = new ArrayList<>();
+ Arrays.sort(candidates); // 先进行排序
+ backtracking(res, new ArrayList<>(), candidates, target, 0, 0);
+ return res;
}
- public void backTracking(int[] arr, int n, int k, int startIndex) {
- //如果 n 小于0,没必要继续本次递归,已经不符合要求了
- if (n < 0) {
+ public void backtracking(List> res, List path, int[] candidates, int target, int sum, int idx) {
+ // 找到了数字和为 target 的组合
+ if (sum == target) {
+ res.add(new ArrayList<>(path));
return;
}
- if (deque.size() == k) {
- if (n == 0) {
- lists.add(new ArrayList(deque));
- }
- return;
- }
- for (int i = startIndex; i < arr.length - (k - deque.size()) + 1; i++) {
- deque.push(arr[i]);
- //减去当前元素
- n -= arr[i];
- backTracking(arr, n, k, i + 1);
- //恢复n
- n += deque.pop();
+
+ for (int i = idx; i < candidates.length; i++) {
+ // 如果 sum + candidates[i] > target 就终止遍历
+ if (sum + candidates[i] > target) break;
+ path.add(candidates[i]);
+ backtracking(res, path, candidates, target, sum + candidates[i], i);
+ path.remove(path.size() - 1); // 回溯,移除路径 path 最后一个元素
}
}
}
From b1b389da3204c4faf0c65519e8f58cf59afe18d4 Mon Sep 17 00:00:00 2001
From: xll <18574553598@163.com>
Date: Sat, 29 May 2021 22:20:52 +0800
Subject: [PATCH 22/95] =?UTF-8?q?=E4=BA=8C=E5=8F=89=E6=A0=91=E8=B7=AF?=
=?UTF-8?q?=E5=BE=84=E6=80=BB=E5=92=8CJavaScript=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0112.路径总和.md | 56 +++++++++++++++++++++++++++++++++++
1 file changed, 56 insertions(+)
diff --git a/problems/0112.路径总和.md b/problems/0112.路径总和.md
index b4a3f38b..4ccd8912 100644
--- a/problems/0112.路径总和.md
+++ b/problems/0112.路径总和.md
@@ -524,6 +524,62 @@ let pathSum = function (root, targetSum) {
};
```
+0112 路径总和
+```javascript
+var hasPathSum = function(root, targetSum) {
+ //递归方法
+ // 1. 确定函数参数
+ const traversal = function(node,count){
+ // 2. 确定终止条件
+ if(node.left===null&&node.right===null&&count===0){
+ return true;
+ }
+ if(node.left===null&&node.right===null){
+ return false;
+ }
+ //3. 单层递归逻辑
+ if(node.left){
+ if(traversal(node.left,count-node.left.val)){
+ return true;
+ }
+ }
+ if(node.right){
+ if(traversal(node.right,count-node.right.val)){
+ return true;
+ }
+ }
+ return false;
+ }
+ if(root===null){
+ return false;
+ }
+ return traversal(root,targetSum-root.val);
+};
+```
+113 路径总和
+```javascript
+var pathSum = function(root, targetSum) {
+ //递归方法
+ let resPath = [],curPath = [];
+ // 1. 确定递归函数参数
+ const travelTree = function(node,count){
+ curPath.push(node.val);
+ count-=node.val;
+ if(node.left===null&&node.right===null&&count===0){
+ resPath.push([...curPath]);
+ }
+ node.left&&travelTree(node.left,count);
+ node.right&&travelTree(node.right,count);
+ let cur = curPath.pop();
+ count-=cur;
+ }
+ if(root===null){
+ return resPath;
+ }
+ travelTree(root,targetSum);
+ return resPath;
+};
+```
From 406d44819ac878232b49a6ffafad06d944643a6a Mon Sep 17 00:00:00 2001
From: X-shuffle <53906918+X-shuffle@users.noreply.github.com>
Date: Sun, 30 May 2021 10:38:30 +0800
Subject: [PATCH 23/95] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=20=E4=BA=8C=E5=8F=89?=
=?UTF-8?q?=E6=A0=91=E7=9A=84=E5=B1=82=E5=BA=8F=E9=81=8D=E5=8E=86=20GO?=
=?UTF-8?q?=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
添加 二叉树的层序遍历 GO版本
---
problems/0102.二叉树的层序遍历.md | 243 ++++++++++++++++++++++
1 file changed, 243 insertions(+)
diff --git a/problems/0102.二叉树的层序遍历.md b/problems/0102.二叉树的层序遍历.md
index ee93911e..00607082 100644
--- a/problems/0102.二叉树的层序遍历.md
+++ b/problems/0102.二叉树的层序遍历.md
@@ -836,6 +836,249 @@ func levelOrder(root *TreeNode) [][]int {
return result
}
```
+> 二叉树的层序遍历(GO语言完全版)
+
+```go
+/**
+102. 二叉树的层序遍历
+ */
+func levelOrder(root *TreeNode) [][]int {
+ res:=[][]int{}
+ if root==nil{//防止为空
+ return res
+ }
+ queue:=list.New()
+ queue.PushBack(root)
+ var tmpArr []int
+ for queue.Len()>0 {
+ length:=queue.Len()//保存当前层的长度,然后处理当前层(十分重要,防止添加下层元素影响判断层中元素的个数)
+ for i:=0;i0{
+ length:=queue.Len()
+ tmp:=[]int{}
+ for i:=0;i0{
+ length:=queue.Len()
+ tmp:=[]int{}
+ for i:=0;i0 {
+ length:=queue.Len()//保存当前层的长度,然后处理当前层(十分重要,防止添加下层元素影响判断层中元素的个数)
+ for i:=0;i0{
+ length:=queue.Len()//记录当前层的数量
+ var tmp []int
+ for T:=0;T0 {
+ length:=queue.Len()//保存当前层的长度,然后处理当前层(十分重要,防止添加下层元素影响判断层中元素的个数)
+ for i:=0;i max {
+ max = val
+ }
+ }
+ return max
+}
+/**
+116. 填充每个节点的下一个右侧节点指针
+117. 填充每个节点的下一个右侧节点指针 II
+ */
+
+func connect(root *Node) *Node {
+ res:=[][]*Node{}
+ if root==nil{//防止为空
+ return root
+ }
+ queue:=list.New()
+ queue.PushBack(root)
+ var tmpArr []*Node
+ for queue.Len()>0 {
+ length:=queue.Len()//保存当前层的长度,然后处理当前层(十分重要,防止添加下层元素影响判断层中元素的个数)
+ for i:=0;i
Date: Sun, 30 May 2021 14:07:55 +0800
Subject: [PATCH 24/95] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200203.=E7=A7=BB?=
=?UTF-8?q?=E9=99=A4=E9=93=BE=E8=A1=A8=E5=85=83=E7=B4=A0=20python3?=
=?UTF-8?q?=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0203.移除链表元素.md | 18 +++++++++++++++++-
1 file changed, 17 insertions(+), 1 deletion(-)
diff --git a/problems/0203.移除链表元素.md b/problems/0203.移除链表元素.md
index dce5d265..cac9f233 100644
--- a/problems/0203.移除链表元素.md
+++ b/problems/0203.移除链表元素.md
@@ -208,7 +208,23 @@ public ListNode removeElements(ListNode head, int val) {
```
Python:
-
+```python
+# Definition for singly-linked list.
+# class ListNode:
+# def __init__(self, val=0, next=None):
+# self.val = val
+# self.next = next
+class Solution:
+ def removeElements(self, head: ListNode, val: int) -> ListNode:
+ dummy_head = ListNode(next=head) #添加一个虚拟节点
+ cur = dummy_head
+ while(cur.next!=None):
+ if(cur.next.val == val):
+ cur.next = cur.next.next #删除cur.next节点
+ else:
+ cur = cur.next
+ return dummy_head.next
+```
Go:
From 5e1860876b8d594d2e4d74a03772495c06cc357f Mon Sep 17 00:00:00 2001
From: evanlai
Date: Sun, 30 May 2021 14:10:11 +0800
Subject: [PATCH 25/95] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200206.=E7=BF=BB?=
=?UTF-8?q?=E8=BD=AC=E9=93=BE=E8=A1=A8=20python3=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0206.翻转链表.md | 20 +++++++++++++++++++-
1 file changed, 19 insertions(+), 1 deletion(-)
diff --git a/problems/0206.翻转链表.md b/problems/0206.翻转链表.md
index 52ef6484..7c002382 100644
--- a/problems/0206.翻转链表.md
+++ b/problems/0206.翻转链表.md
@@ -143,7 +143,25 @@ class Solution {
```
Python:
-
+```python
+#双指针
+# Definition for singly-linked list.
+# class ListNode:
+# def __init__(self, val=0, next=None):
+# self.val = val
+# self.next = next
+class Solution:
+ def reverseList(self, head: ListNode) -> ListNode:
+ cur = head
+ pre = None
+ while(cur!=None):
+ temp = cur.next # 保存一下 cur的下一个节点,因为接下来要改变cur->next
+ cur.next = pre #反转
+ #更新pre、cur指针
+ pre = cur
+ cur = temp
+ return pre
+```
Go:
From 013ba0575ed1729e8a1c31c251313e379d38adc2 Mon Sep 17 00:00:00 2001
From: evanlai
Date: Sun, 30 May 2021 14:13:38 +0800
Subject: [PATCH 26/95] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200019.=E5=88=A0?=
=?UTF-8?q?=E9=99=A4=E9=93=BE=E8=A1=A8=E7=9A=84=E5=80=92=E6=95=B0=E7=AC=AC?=
=?UTF-8?q?N=E4=B8=AA=E8=8A=82=E7=82=B9=20python3=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
...0019.删除链表的倒数第N个节点.md | 24 +++++++++++++++++++
1 file changed, 24 insertions(+)
diff --git a/problems/0019.删除链表的倒数第N个节点.md b/problems/0019.删除链表的倒数第N个节点.md
index 7d2fe97e..52735794 100644
--- a/problems/0019.删除链表的倒数第N个节点.md
+++ b/problems/0019.删除链表的倒数第N个节点.md
@@ -112,6 +112,30 @@ class Solution {
}
}
```
+
+Python:
+```python
+# Definition for singly-linked list.
+# class ListNode:
+# def __init__(self, val=0, next=None):
+# self.val = val
+# self.next = next
+class Solution:
+ def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
+ head_dummy = ListNode()
+ head_dummy.next = head
+
+ slow, fast = head_dummy, head_dummy
+ while(n!=0): #fast先往前走n步
+ fast = fast.next
+ n -= 1
+ while(fast.next!=None):
+ slow = slow.next
+ fast = fast.next
+ #fast 走到结尾后,slow的下一个节点为倒数第N个节点
+ slow.next = slow.next.next #删除
+ return head_dummy.next
+```
Go:
```Go
/**
From e98fb5a56f256371d01ed7e9456aa0fde805626f Mon Sep 17 00:00:00 2001
From: evanlai
Date: Sun, 30 May 2021 14:21:41 +0800
Subject: [PATCH 27/95] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=20=E9=9D=A2=E8=AF=95?=
=?UTF-8?q?=E9=A2=9802.07.=E9=93=BE=E8=A1=A8=E7=9B=B8=E4=BA=A4=20python3?=
=?UTF-8?q?=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/面试题02.07.链表相交.md | 37 +++++++++++++++++++++++++
1 file changed, 37 insertions(+)
diff --git a/problems/面试题02.07.链表相交.md b/problems/面试题02.07.链表相交.md
index 13014dd1..78f34e71 100644
--- a/problems/面试题02.07.链表相交.md
+++ b/problems/面试题02.07.链表相交.md
@@ -151,7 +151,44 @@ public class Solution {
```
Python:
+```python
+# Definition for singly-linked list.
+# class ListNode:
+# def __init__(self, x):
+# self.val = x
+# self.next = None
+class Solution:
+ def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode:
+ lengthA,lengthB = 0,0
+ curA,curB = headA,headB
+ while(curA!=None): #求链表A的长度
+ curA = curA.next
+ lengthA +=1
+
+ while(curB!=None): #求链表B的长度
+ curB = curB.next
+ lengthB +=1
+
+ curA, curB = headA, headB
+
+ if lengthB>lengthA: #让curA为最长链表的头,lenA为其长度
+ lengthA, lengthB = lengthB, lengthA
+ curA, curB = curB, curA
+
+ gap = lengthA - lengthB #求长度差
+ while(gap!=0):
+ curA = curA.next #让curA和curB在同一起点上
+ gap -= 1
+
+ while(curA!=None):
+ if curA == curB:
+ return curA
+ else:
+ curA = curA.next
+ curB = curB.next
+ return None
+```
Go:
From bf2ffec6674a429663a11aca03abbd9f36a9864c Mon Sep 17 00:00:00 2001
From: LiangDazhu <42199191+LiangDazhu@users.noreply.github.com>
Date: Sun, 30 May 2021 19:21:51 +0800
Subject: [PATCH 28/95] =?UTF-8?q?Update=200416.=E5=88=86=E5=89=B2=E7=AD=89?=
=?UTF-8?q?=E5=92=8C=E5=AD=90=E9=9B=86.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
Added python version code
---
problems/0416.分割等和子集.md | 14 ++++++++++++--
1 file changed, 12 insertions(+), 2 deletions(-)
diff --git a/problems/0416.分割等和子集.md b/problems/0416.分割等和子集.md
index e69eb1a4..55d200e2 100644
--- a/problems/0416.分割等和子集.md
+++ b/problems/0416.分割等和子集.md
@@ -222,8 +222,18 @@ class Solution {
```
Python:
-
-
+```python
+class Solution:
+ def canPartition(self, nums: List[int]) -> bool:
+ taraget = sum(nums)
+ if taraget % 2 == 1: return False
+ taraget //= 2
+ dp = [0] * 10001
+ for i in range(len(nums)):
+ for j in range(taraget, nums[i] - 1, -1):
+ dp[j] = max(dp[j], dp[j - nums[i]] + nums[i])
+ return taraget == dp[taraget]
+```
Go:
From 1516dc3d152b818f1e3d4f3675924c35918e0558 Mon Sep 17 00:00:00 2001
From: resyon
Date: Sun, 30 May 2021 21:27:11 +0800
Subject: [PATCH 29/95] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=E4=BA=860239-golang?=
=?UTF-8?q?=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0239.滑动窗口最大值.md | 30 ++++++++++++++++++++++++++
1 file changed, 30 insertions(+)
diff --git a/problems/0239.滑动窗口最大值.md b/problems/0239.滑动窗口最大值.md
index 6cc355ca..ed17157a 100644
--- a/problems/0239.滑动窗口最大值.md
+++ b/problems/0239.滑动窗口最大值.md
@@ -267,6 +267,36 @@ Python:
Go:
+```go
+func maxSlidingWindow(nums []int, k int) []int {
+ var queue []int
+ var rtn []int
+
+ for f := 0; f < len(nums); f++ {
+ //维持队列递减, 将 k 插入合适的位置, queue中 <=k 的 元素都不可能是窗口中的最大值, 直接弹出
+ for len(queue) > 0 && nums[f] > nums[queue[len(queue)-1]] {
+ queue = queue[:len(queue)-1]
+ }
+ // 等大的后来者也应入队
+ if len(queue) == 0 || nums[f] <= nums[queue[len(queue)-1]] {
+ queue = append(queue, f)
+ }
+
+ if f >= k - 1 {
+ rtn = append(rtn, nums[queue[0]])
+ //弹出离开窗口的队首
+ if f - k + 1 == queue[0] {
+ queue = queue[1:]
+ }
+ }
+ }
+
+ return rtn
+
+}
+
+```
+
Javascript:
```javascript
var maxSlidingWindow = function (nums, k) {
From 0de459b1ff5f491e13c4fec93cbc7727dabc73e6 Mon Sep 17 00:00:00 2001
From: LiangDazhu <42199191+LiangDazhu@users.noreply.github.com>
Date: Sun, 30 May 2021 22:54:01 +0800
Subject: [PATCH 30/95] =?UTF-8?q?Update=20=E8=83=8C=E5=8C=85=E7=90=86?=
=?UTF-8?q?=E8=AE=BA=E5=9F=BA=E7=A1=8001=E8=83=8C=E5=8C=85-2.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
dp[j]可以通过dp[j - weight[i]]推导出来, 应该是写错了吧
---
problems/背包理论基础01背包-2.md | 2 +-
1 file changed, 1 insertion(+), 1 deletion(-)
diff --git a/problems/背包理论基础01背包-2.md b/problems/背包理论基础01背包-2.md
index a169425a..e831d88f 100644
--- a/problems/背包理论基础01背包-2.md
+++ b/problems/背包理论基础01背包-2.md
@@ -55,7 +55,7 @@
dp[j]为 容量为j的背包所背的最大价值,那么如何推导dp[j]呢?
-dp[j]可以通过dp[j - weight[j]]推导出来,dp[j - weight[i]]表示容量为j - weight[i]的背包所背的最大价值。
+dp[j]可以通过dp[j - weight[i]]推导出来,dp[j - weight[i]]表示容量为j - weight[i]的背包所背的最大价值。
dp[j - weight[i]] + value[i] 表示 容量为 j - 物品i重量 的背包 加上 物品i的价值。(也就是容量为j的背包,放入物品i了之后的价值即:dp[j])
From 60ad3b081134174577ad58416d275b1bb78267cd Mon Sep 17 00:00:00 2001
From: fusunx <1102654482@qq.com>
Date: Sun, 30 May 2021 23:37:15 +0800
Subject: [PATCH 31/95] =?UTF-8?q?0045.=E8=B7=B3=E8=B7=83=E6=B8=B8=E6=88=8F?=
=?UTF-8?q?||.md=20Javascript?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0045.跳跃游戏II.md | 20 ++++++++++++++++++++
1 file changed, 20 insertions(+)
diff --git a/problems/0045.跳跃游戏II.md b/problems/0045.跳跃游戏II.md
index 31b52b31..4128da4c 100644
--- a/problems/0045.跳跃游戏II.md
+++ b/problems/0045.跳跃游戏II.md
@@ -208,6 +208,26 @@ func jump(nums []int) int {
}
return dp[len(nums)-1]
}
+```
+
+Javascript:
+```Javascript
+var jump = function(nums) {
+ let curIndex = 0
+ let nextIndex = 0
+ let steps = 0
+ for(let i = 0; i < nums.length - 1; i++) {
+ nextIndex = Math.max(nums[i] + i, nextIndex)
+ if(i === curIndex) {
+ curIndex = nextIndex
+ steps++
+ }
+ }
+
+ return steps
+};
+```
+
/*
dp[i]表示从起点到当前位置的最小跳跃次数
dp[i]=min(dp[j]+1,dp[i]) 表示从j位置用一步跳跃到当前位置,这个j位置可能有很多个,却最小一个就可以
From e9c91e78998bde1a123865df30aca4071bba8159 Mon Sep 17 00:00:00 2001
From: "qingyi.liu"
Date: Mon, 31 May 2021 10:55:43 +0800
Subject: [PATCH 32/95] =?UTF-8?q?=E4=BA=8C=E5=8F=89=E6=A0=91=E7=9A=84?=
=?UTF-8?q?=E7=BB=9F=E4=B8=80=E8=BF=AD=E4=BB=A3=E6=B3=95=20javaScript?=
=?UTF-8?q?=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/二叉树的统一迭代法.md | 80 +++++++++++++++++++++++++
1 file changed, 80 insertions(+)
diff --git a/problems/二叉树的统一迭代法.md b/problems/二叉树的统一迭代法.md
index dca5d3e3..f4091ad5 100644
--- a/problems/二叉树的统一迭代法.md
+++ b/problems/二叉树的统一迭代法.md
@@ -374,6 +374,86 @@ func postorderTraversal(root *TreeNode) []int {
}
```
+javaScript:
+
+> 前序遍历统一迭代法
+
+```js
+
+// 前序遍历:中左右
+// 压栈顺序:右左中
+
+var preorderTraversal = function(root, res = []) {
+ const stack = [];
+ if (root) stack.push(root);
+ while(stack.length) {
+ const node = stack.pop();
+ if(!node) {
+ res.push(stack.pop().val);
+ continue;
+ }
+ if (node.right) stack.push(node.right); // 右
+ if (node.left) stack.push(node.left); // 左
+ stack.push(node); // 中
+ stack.push(null);
+ };
+ return res;
+};
+
+```
+
+> 中序遍历统一迭代法
+
+```js
+
+// 中序遍历:左中右
+// 压栈顺序:右中左
+
+var inorderTraversal = function(root, res = []) {
+ const stack = [];
+ if (root) stack.push(root);
+ while(stack.length) {
+ const node = stack.pop();
+ if(!node) {
+ res.push(stack.pop().val);
+ continue;
+ }
+ if (node.right) stack.push(node.right); // 右
+ stack.push(node); // 中
+ stack.push(null);
+ if (node.left) stack.push(node.left); // 左
+ };
+ return res;
+};
+
+```
+
+> 后序遍历统一迭代法
+
+```js
+
+// 后续遍历:左右中
+// 压栈顺序:中右左
+
+var postorderTraversal = function(root, res = []) {
+ const stack = [];
+ if (root) stack.push(root);
+ while(stack.length) {
+ const node = stack.pop();
+ if(!node) {
+ res.push(stack.pop().val);
+ continue;
+ }
+ stack.push(node); // 中
+ stack.push(null);
+ if (node.right) stack.push(node.right); // 右
+ if (node.left) stack.push(node.left); // 左
+ };
+ return res;
+};
+
+```
+
-----------------------
From 18a37ec17d2e596d12f37373b0decffded380f76 Mon Sep 17 00:00:00 2001
From: "qingyi.liu"
Date: Mon, 31 May 2021 15:25:34 +0800
Subject: [PATCH 33/95] =?UTF-8?q?=E4=BA=8C=E5=8F=89=E6=A0=91=E7=9A=84?=
=?UTF-8?q?=E5=B1=82=E5=BA=8F=E9=81=8D=E5=8E=86=20=E4=B8=80=E5=A5=97?=
=?UTF-8?q?=E6=89=93=E5=85=AB=E4=B8=AAJavaScript=20=E8=BF=AD=E4=BB=A3=20+?=
=?UTF-8?q?=20=E9=80=92=E5=BD=92=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0102.二叉树的层序遍历.md | 270 ++++++++++++++++++++++
1 file changed, 270 insertions(+)
diff --git a/problems/0102.二叉树的层序遍历.md b/problems/0102.二叉树的层序遍历.md
index 00607082..2be38bc4 100644
--- a/problems/0102.二叉树的层序遍历.md
+++ b/problems/0102.二叉树的层序遍历.md
@@ -1100,6 +1100,276 @@ var levelOrder = function (root) {
};
```
+> 二叉树的层序遍历(Javascript语言完全版) (迭代 + 递归)
+
+```js
+/**
+ * 102. 二叉树的层序遍历
+ * @param {TreeNode} root
+ * @return {number[][]}
+ */
+
+// 迭代
+
+var levelOrder = function(root) {
+ const queue = [], res = [];
+ root && queue.push(root);
+ while(len = queue.length) {
+ const val = [];
+ while(len--) {
+ const node = queue.shift();
+ val.push(node.val);
+ node.left && queue.push(node.left);
+ node.right && queue.push(node.right);
+ }
+ res.push(val);
+ }
+ return res;
+};
+
+// 递归
+var levelOrder = function(root) {
+ const res = [];
+ function defs (root, i) {
+ if(!root) return;
+ if(!res[i]) res[i] = [];
+ res[i].push(root.val)
+ root.left && defs(root.left, i + 1);
+ root.right && defs(root.right, i + 1);
+ }
+ defs(root, 0);
+ return res;
+};
+
+
+/**
+ * 107. 二叉树的层序遍历 II
+ * @param {TreeNode} root
+ * @return {number[][]}
+ */
+
+// 迭代
+
+var levelOrderBottom = function(root) {
+ const queue = [], res = [];
+ root && queue.push(root);
+ while(len = queue.length) {
+ const val = [];
+ while(len--) {
+ const node = queue.shift();
+ val.push(node.val);
+ node.left && queue.push(node.left);
+ node.right && queue.push(node.right);
+ }
+ res.push(val);
+ }
+ return res.reverse()
+};
+
+// 递归
+
+var levelOrderBottom = function(root) {
+ const res = [];
+ function defs (root, i) {
+ if(!root) return;
+ if(!res[i]) res[i] = [];
+ res[i].push(root.val);
+ root.left && defs(root.left, i + 1);
+ root.right && defs(root.right, i + 1);
+ }
+ defs(root, 0);
+ return res.reverse();
+};
+
+/**
+ * 199. 二叉树的右视图
+ * @param {TreeNode} root
+ * @return {number[]}
+ */
+
+// 迭代
+
+var rightSideView = function(root) {
+ const res = [], queue = [];
+ root && queue.push(root);
+ while(l = queue.length) {
+ while (l--) {
+ const {val, left, right} = queue.shift();
+ !l && res.push(val);
+ left && queue.push(left);
+ right && queue.push(right);
+ }
+ }
+ return res;
+};
+
+// 递归
+var rightSideView = function(root) {
+ const res = [];
+ function defs(root, i) {
+ if(!root) return;
+ res[i] = root.val;
+ root.left && defs(root.left, i + 1);
+ root.right && defs(root.right, i + 1);
+ }
+ defs(root, 0);
+ return res;
+};
+
+/**
+ * 637. 二叉树的层平均值
+ * @param {TreeNode} root
+ * @return {number[]}
+ */
+
+// 迭代
+var averageOfLevels = function(root) {
+ const stack = [], res = [];
+ root && stack.push(root);
+ while(len = stack.length) {
+ let sum = 0, l = len;
+ while(l--) {
+ const {val, left, right} = stack.shift();
+ sum += val;
+ left && stack.push(left);
+ right && stack.push(right);
+ }
+ res.push(sum/len);
+ }
+ return res;
+};
+
+// 递归
+var averageOfLevels = function(root) {
+ const resCount = [], res = [];
+ function defs(root, i) {
+ if(!root) return;
+ if(isNaN(res[i])) resCount[i] = res[i] = 0;
+ res[i] += root.val;
+ resCount[i]++;
+ root.left && defs(root.left, i + 1);
+ root.right && defs(root.right, i + 1);
+ }
+ defs(root, 0);
+ return res.map((val, i) => val / resCount[i]);
+};
+
+/**
+ * 515. 在每个树行中找最大值
+ * @param {TreeNode} root
+ * @return {number[]}
+ */
+
+// 迭代
+const MIN_G = Number.MIN_SAFE_INTEGER;
+var largestValues = function(root) {
+ const stack = [], res = [];
+ root && stack.push(root);
+ while(len = stack.length) {
+ let max = MIN_G;
+ while(len--) {
+ const {val, left, right} = stack.shift();
+ max = max > val ? max : val;
+ left && stack.push(left);
+ right && stack.push(right);
+ }
+ res.push(max);
+ }
+ return res;
+};
+
+// 递归
+var largestValues = function(root) {
+ const res = [];
+ function defs (root, i) {
+ if(!root) return;
+ if(isNaN(res[i])) res[i] = root.val;
+ res[i] = res[i] > root.val ? res[i] : root.val;
+ root.left && defs(root.left, i + 1);
+ root.right && defs(root.right, i + 1);
+ }
+ defs(root, 0);
+ return res;
+};
+
+/**
+ * 429. N 叉树的层序遍历
+ * @param {Node|null} root
+ * @return {number[][]}
+ */
+
+// 迭代
+var levelOrder = function(root) {
+ const stack = [], res = [];
+ root && stack.push(root);
+ while(len = stack.length) {
+ const vals = [];
+ while(len--) {
+ const {val, children} = stack.shift();
+ vals.push(val);
+ for(const e of children) {
+ stack.push(e);
+ }
+ }
+ res.push(vals);
+ }
+ return res;
+};
+
+// 递归
+
+var levelOrder = function(root) {
+ const res = [];
+ function defs (root, i) {
+ if(!root) return;
+ if(!res[i]) res[i] = [];
+ res[i].push(root.val);
+ for(const e of root.children) {
+ defs(e, i + 1);
+ }
+ }
+ defs(root, 0);
+ return res;
+};
+
+/**
+ * 116. 填充每个节点的下一个右侧节点指针
+ * 117. 填充每个节点的下一个右侧节点指针 II
+ * @param {Node} root
+ * @return {Node}
+ */
+
+// 迭代
+var connect = function(root) {
+ const stack = [];
+ root && stack.push(root);
+ while(len = stack.length) {
+ while(len--) {
+ const node1 = stack.shift(),
+ node2 = len ? stack[0] : null;
+ node1.next = node2;
+ node1.left && stack.push(node1.left);
+ node1.right && stack.push(node1.right);
+ }
+ }
+ return root;
+};
+
+// 递归
+var connect = function(root) {
+ const res = [];
+ function defs (root, i) {
+ if(!root) return;
+ if(res[i]) res[i].next = root;
+ res[i] = root;
+ root.left && defs(root.left, i + 1);
+ root.right && defs(root.right, i + 1);
+ }
+ defs(root, 0);
+ return root;
+};
+```
+
-----------------------
* 作者微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
* B站视频:[代码随想录](https://space.bilibili.com/525438321)
From 4a5c0cb75b1c265f2dacfb3826dccdb3f0bdae98 Mon Sep 17 00:00:00 2001
From: "qingyi.liu"
Date: Mon, 31 May 2021 17:19:02 +0800
Subject: [PATCH 34/95] update: stack to queue
---
problems/0102.二叉树的层序遍历.md | 48 +++++++++++------------
1 file changed, 24 insertions(+), 24 deletions(-)
diff --git a/problems/0102.二叉树的层序遍历.md b/problems/0102.二叉树的层序遍历.md
index 2be38bc4..51bd8510 100644
--- a/problems/0102.二叉树的层序遍历.md
+++ b/problems/0102.二叉树的层序遍历.md
@@ -1224,15 +1224,15 @@ var rightSideView = function(root) {
// 迭代
var averageOfLevels = function(root) {
- const stack = [], res = [];
- root && stack.push(root);
- while(len = stack.length) {
+ const queue = [], res = [];
+ root && queue.push(root);
+ while(len = queue.length) {
let sum = 0, l = len;
while(l--) {
- const {val, left, right} = stack.shift();
+ const {val, left, right} = queue.shift();
sum += val;
- left && stack.push(left);
- right && stack.push(right);
+ left && queue.push(left);
+ right && queue.push(right);
}
res.push(sum/len);
}
@@ -1263,15 +1263,15 @@ var averageOfLevels = function(root) {
// 迭代
const MIN_G = Number.MIN_SAFE_INTEGER;
var largestValues = function(root) {
- const stack = [], res = [];
- root && stack.push(root);
- while(len = stack.length) {
+ const queue = [], res = [];
+ root && queue.push(root);
+ while(len = queue.length) {
let max = MIN_G;
while(len--) {
- const {val, left, right} = stack.shift();
+ const {val, left, right} = queue.shift();
max = max > val ? max : val;
- left && stack.push(left);
- right && stack.push(right);
+ left && queue.push(left);
+ right && queue.push(right);
}
res.push(max);
}
@@ -1300,15 +1300,15 @@ var largestValues = function(root) {
// 迭代
var levelOrder = function(root) {
- const stack = [], res = [];
- root && stack.push(root);
- while(len = stack.length) {
+ const queue = [], res = [];
+ root && queue.push(root);
+ while(len = queue.length) {
const vals = [];
while(len--) {
- const {val, children} = stack.shift();
+ const {val, children} = queue.shift();
vals.push(val);
for(const e of children) {
- stack.push(e);
+ queue.push(e);
}
}
res.push(vals);
@@ -1341,15 +1341,15 @@ var levelOrder = function(root) {
// 迭代
var connect = function(root) {
- const stack = [];
- root && stack.push(root);
- while(len = stack.length) {
+ const queue = [];
+ root && queue.push(root);
+ while(len = queue.length) {
while(len--) {
- const node1 = stack.shift(),
- node2 = len ? stack[0] : null;
+ const node1 = queue.shift(),
+ node2 = len ? queue[0] : null;
node1.next = node2;
- node1.left && stack.push(node1.left);
- node1.right && stack.push(node1.right);
+ node1.left && queue.push(node1.left);
+ node1.right && queue.push(node1.right);
}
}
return root;
From 17b8c2d9acb2481783b7820717a0db989c0dd961 Mon Sep 17 00:00:00 2001
From: Lulu
Date: Mon, 31 May 2021 20:37:11 +0800
Subject: [PATCH 35/95] fix typo
---
.../关于时间复杂度,你不知道的都在这里!.md | 2 +-
1 file changed, 1 insertion(+), 1 deletion(-)
diff --git a/problems/前序/关于时间复杂度,你不知道的都在这里!.md b/problems/前序/关于时间复杂度,你不知道的都在这里!.md
index bd3bd284..3f5bc156 100644
--- a/problems/前序/关于时间复杂度,你不知道的都在这里!.md
+++ b/problems/前序/关于时间复杂度,你不知道的都在这里!.md
@@ -117,7 +117,7 @@ O(2 * n^2 + 10 * n + 1000) < O(3 * n^2),所以说最后省略掉常数项系

-假如有两个算法的时间复杂度,分别是log以2为底n的对数和log以10为底n的对数,那么这里如果还记得高中数学的话,应该不能理解`以2为底n的对数 = 以2为底10的对数 * 以10为底n的对数`。
+假如有两个算法的时间复杂度,分别是log以2为底n的对数和log以10为底n的对数,那么这里如果还记得高中数学的话,应该不难理解`以2为底n的对数 = 以2为底10的对数 * 以10为底n的对数`。
而以2为底10的对数是一个常数,在上文已经讲述了我们计算时间复杂度是忽略常数项系数的。
From 276a7b1e14ca7a54ac9826bde248e0a1c6833444 Mon Sep 17 00:00:00 2001
From: LiangDazhu <42199191+LiangDazhu@users.noreply.github.com>
Date: Mon, 31 May 2021 21:34:11 +0800
Subject: [PATCH 36/95] =?UTF-8?q?Update=201049.=E6=9C=80=E5=90=8E=E4=B8=80?=
=?UTF-8?q?=E5=9D=97=E7=9F=B3=E5=A4=B4=E7=9A=84=E9=87=8D=E9=87=8FII.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
Added python version code
---
problems/1049.最后一块石头的重量II.md | 12 +++++++++++-
1 file changed, 11 insertions(+), 1 deletion(-)
diff --git a/problems/1049.最后一块石头的重量II.md b/problems/1049.最后一块石头的重量II.md
index fcd3712a..f74600b1 100644
--- a/problems/1049.最后一块石头的重量II.md
+++ b/problems/1049.最后一块石头的重量II.md
@@ -178,7 +178,17 @@ class Solution {
```
Python:
-
+```python
+class Solution:
+ def lastStoneWeightII(self, stones: List[int]) -> int:
+ sumweight = sum(stones)
+ target = sumweight // 2
+ dp = [0] * 15001
+ for i in range(len(stones)):
+ for j in range(target, stones[i] - 1, -1):
+ dp[j] = max(dp[j], dp[j - stones[i]] + stones[i])
+ return sumweight - 2 * dp[target]
+```
Go:
From 659595f2b4d09d45d7301b728ac274b1cd52b39a Mon Sep 17 00:00:00 2001
From: NevS <1173325467@qq.com>
Date: Mon, 31 May 2021 22:26:44 +0800
Subject: [PATCH 37/95] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200202.=E5=BF=AB?=
=?UTF-8?q?=E4=B9=90=E6=95=B0=20go=E7=89=88?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0202.快乐数.md | 18 ++++++++++++++++++
1 file changed, 18 insertions(+)
diff --git a/problems/0202.快乐数.md b/problems/0202.快乐数.md
index c07405ec..396231fe 100644
--- a/problems/0202.快乐数.md
+++ b/problems/0202.快乐数.md
@@ -111,6 +111,24 @@ Python:
Go:
+```go
+func isHappy(n int) bool {
+ m := make(map[int]bool)
+ for n != 1 && !m[n] {
+ n, m[n] = getSum(n), true
+ }
+ return n == 1
+}
+
+func getSum(n int) int {
+ sum := 0
+ for n > 0 {
+ sum += (n % 10) * (n % 10)
+ n = n / 10
+ }
+ return sum
+}
+```
javaScript:
From 605b9159de96de727cf4592144783c6d203f6fbe Mon Sep 17 00:00:00 2001
From: LiangDazhu <42199191+LiangDazhu@users.noreply.github.com>
Date: Mon, 31 May 2021 22:49:40 +0800
Subject: [PATCH 38/95] =?UTF-8?q?Update=200494.=E7=9B=AE=E6=A0=87=E5=92=8C?=
=?UTF-8?q?.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
Added python version code
---
problems/0494.目标和.md | 16 ++++++++++++++--
1 file changed, 14 insertions(+), 2 deletions(-)
diff --git a/problems/0494.目标和.md b/problems/0494.目标和.md
index a9ddc768..c772a09d 100644
--- a/problems/0494.目标和.md
+++ b/problems/0494.目标和.md
@@ -150,7 +150,7 @@ dp[j] 表示:填满j(包括j)这么大容积的包,有dp[i]种方法
有哪些来源可以推出dp[j]呢?
-不考虑nums[i]的情况下,填满容量为j - nums[i]的背包,有dp[j - nums[i]]中方法。
+不考虑nums[i]的情况下,填满容量为j - nums[i]的背包,有dp[j - nums[i]]种方法。
那么只要搞到nums[i]的话,凑成dp[j]就有dp[j - nums[i]] 种方法。
@@ -261,7 +261,19 @@ class Solution {
```
Python:
-
+```python
+class Solution:
+ def findTargetSumWays(self, nums: List[int], target: int) -> int:
+ sumValue = sum(nums)
+ if target > sumValue or (sumValue + target) % 2 == 1: return 0
+ bagSize = (sumValue + target) // 2
+ dp = [0] * (bagSize + 1)
+ dp[0] = 1
+ for i in range(len(nums)):
+ for j in range(bagSize, nums[i] - 1, -1):
+ dp[j] += dp[j - nums[i]]
+ return dp[bagSize]
+```
Go:
From 425f407d6c66854446a1a2654bc1639171aee172 Mon Sep 17 00:00:00 2001
From: Baturu <45113401+z80160280@users.noreply.github.com>
Date: Mon, 31 May 2021 11:20:40 -0700
Subject: [PATCH 39/95] =?UTF-8?q?Update=200024.=E4=B8=A4=E4=B8=A4=E4=BA=A4?=
=?UTF-8?q?=E6=8D=A2=E9=93=BE=E8=A1=A8=E4=B8=AD=E7=9A=84=E8=8A=82=E7=82=B9?=
=?UTF-8?q?.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0024.两两交换链表中的节点.md | 17 +++++++++++++++++
1 file changed, 17 insertions(+)
diff --git a/problems/0024.两两交换链表中的节点.md b/problems/0024.两两交换链表中的节点.md
index 643f6055..59ded523 100644
--- a/problems/0024.两两交换链表中的节点.md
+++ b/problems/0024.两两交换链表中的节点.md
@@ -129,6 +129,23 @@ class Solution {
```
Python:
+```python
+class Solution:
+ def swapPairs(self, head: ListNode) -> ListNode:
+ dummy = ListNode(0) #设置一个虚拟头结点
+ dummy.next = head
+ cur = dummy
+ while cur.next and cur.next.next:
+ tmp = cur.next #记录临时节点
+ tmp1 = cur.next.next.next #记录临时节点
+
+ cur.next = cur.next.next #步骤一
+ cur.next.next = tmp #步骤二
+ cur.next.next.next = tmp1 #步骤三
+
+ cur = cur.next.next #cur移动两位,准备下一轮交换
+ return dummy.next
+```
Go:
```go
From e5b81df01cdc6588710edb4bf7a88de6cc8a8da8 Mon Sep 17 00:00:00 2001
From: Yang
Date: Mon, 31 May 2021 16:43:45 -0400
Subject: [PATCH 40/95] =?UTF-8?q?Update=200452.=E7=94=A8=E6=9C=80=E5=B0=91?=
=?UTF-8?q?=E6=95=B0=E9=87=8F=E7=9A=84=E7=AE=AD=E5=BC=95=E7=88=86=E6=B0=94?=
=?UTF-8?q?=E7=90=83.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
Java code:
- Fix an error in sorting (simplified as well)
- Add checking for boundary condition
---
problems/0452.用最少数量的箭引爆气球.md | 12 ++----------
1 file changed, 2 insertions(+), 10 deletions(-)
diff --git a/problems/0452.用最少数量的箭引爆气球.md b/problems/0452.用最少数量的箭引爆气球.md
index a8b55ca9..9b0cc925 100644
--- a/problems/0452.用最少数量的箭引爆气球.md
+++ b/problems/0452.用最少数量的箭引爆气球.md
@@ -142,16 +142,8 @@ Java:
```java
class Solution {
public int findMinArrowShots(int[][] points) {
- Arrays.sort(points, new Comparator() {
- @Override
- public int compare(int[] o1, int[] o2) {
- if (o1[0] != o2[0]) {
- return Integer.compare(o1[0],o2[0]);
- } else {
- return Integer.compare(o1[0],o2[0]);
- }
- }
- });
+ if (points.length == 0) return 0;
+ Arrays.sort(points, (o1, o2) -> Integer.compare(o1[0], o2[0]));
int count = 1;
for (int i = 1; i < points.length; i++) {
From 72b2521873136f8f45648c185fd463f976f586ca Mon Sep 17 00:00:00 2001
From: fusunx <1102654482@qq.com>
Date: Tue, 1 Jun 2021 08:15:47 +0800
Subject: [PATCH 41/95] =?UTF-8?q?1005.K=E6=AC=A1=E5=8F=96=E5=8F=8D.md=20Ja?=
=?UTF-8?q?vascript?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
...1005.K次取反后最大化的数组和.md | 22 +++++++++++++++++++
1 file changed, 22 insertions(+)
diff --git a/problems/1005.K次取反后最大化的数组和.md b/problems/1005.K次取反后最大化的数组和.md
index 43282f25..e57e26ad 100644
--- a/problems/1005.K次取反后最大化的数组和.md
+++ b/problems/1005.K次取反后最大化的数组和.md
@@ -140,7 +140,29 @@ class Solution:
Go:
+Javascript:
+```Javascript
+var largestSumAfterKNegations = function(nums, k) {
+ nums.sort((a, b) => {
+ return Math.abs(b) - Math.abs(a)
+ })
+ for(let i = 0; i < nums.length; i++) {
+ if(nums[i] < 0 && k > 0) {
+ nums[i] *= -1
+ k--
+ }
+ }
+ if(k > 0 && k % 2 === 1) {
+ nums[nums.length - 1] *= -1
+ }
+ k = 0
+
+ return nums.reduce((a, b) => {
+ return a + b
+ })
+};
+```
-----------------------
* 作者微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
From 7a0d134d6e96541deb2fe4757743aed4c3ecd94e Mon Sep 17 00:00:00 2001
From: Baturu <45113401+z80160280@users.noreply.github.com>
Date: Mon, 31 May 2021 18:31:10 -0700
Subject: [PATCH 42/95] =?UTF-8?q?Update=200242.=E6=9C=89=E6=95=88=E7=9A=84?=
=?UTF-8?q?=E5=AD=97=E6=AF=8D=E5=BC=82=E4=BD=8D=E8=AF=8D.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0242.有效的字母异位词.md | 17 ++++++++++++++++-
1 file changed, 16 insertions(+), 1 deletion(-)
diff --git a/problems/0242.有效的字母异位词.md b/problems/0242.有效的字母异位词.md
index 6a6faf63..ba942f70 100644
--- a/problems/0242.有效的字母异位词.md
+++ b/problems/0242.有效的字母异位词.md
@@ -113,7 +113,22 @@ class Solution {
```
Python:
-
+```python
+class Solution:
+ def isAnagram(self, s: str, t: str) -> bool:
+ record = [0] * 26
+ for i in range(len(s)):
+ #并不需要记住字符a的ASCII,只要求出一个相对数值就可以了
+ record[ord(s[i]) - ord("a")] += 1
+ print(record)
+ for i in range(len(t)):
+ record[ord(t[i]) - ord("a")] -= 1
+ for i in range(26):
+ if record[i] != 0:
+ #record数组如果有的元素不为零0,说明字符串s和t 一定是谁多了字符或者谁少了字符。
+ return False
+ return True
+```
Go:
```go
From 4fc751affb0fba096c9263256555d59aa82615aa Mon Sep 17 00:00:00 2001
From: Baturu <45113401+z80160280@users.noreply.github.com>
Date: Mon, 31 May 2021 18:42:19 -0700
Subject: [PATCH 43/95] =?UTF-8?q?Update=200202.=E5=BF=AB=E4=B9=90=E6=95=B0?=
=?UTF-8?q?.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0202.快乐数.md | 24 +++++++++++++++++++++++-
1 file changed, 23 insertions(+), 1 deletion(-)
diff --git a/problems/0202.快乐数.md b/problems/0202.快乐数.md
index c07405ec..999e7d52 100644
--- a/problems/0202.快乐数.md
+++ b/problems/0202.快乐数.md
@@ -108,7 +108,29 @@ class Solution {
```
Python:
-
+```python
+class Solution:
+ def isHappy(self, n: int) -> bool:
+ set_ = set()
+ while 1:
+ sum_ = self.getSum(n)
+ if sum_ == 1:
+ return True
+ #如果这个sum曾经出现过,说明已经陷入了无限循环了,立刻return false
+ if sum_ in set_:
+ return False
+ else:
+ set_.add(sum_)
+ n = sum_
+
+ #取数值各个位上的单数之和
+ def getSum(self, n):
+ sum_ = 0
+ while n > 0:
+ sum_ += (n%10) * (n%10)
+ n //= 10
+ return sum_
+```
Go:
From be0ef14cef18a6aeae95872c82dca0221d16f76b Mon Sep 17 00:00:00 2001
From: "qingyi.liu"
Date: Tue, 1 Jun 2021 12:01:24 +0800
Subject: [PATCH 44/95] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=20101.=20=E5=AF=B9?=
=?UTF-8?q?=E7=A7=B0=E4=BA=8C=E5=8F=89=E6=A0=91=20go=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0101.对称二叉树.md | 48 ++++++++++++++++++++++++++++++++
1 file changed, 48 insertions(+)
diff --git a/problems/0101.对称二叉树.md b/problems/0101.对称二叉树.md
index d8797d30..b58cef2e 100644
--- a/problems/0101.对称二叉树.md
+++ b/problems/0101.对称二叉树.md
@@ -363,6 +363,54 @@ Python:
Go:
+```go
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ * Val int
+ * Left *TreeNode
+ * Right *TreeNode
+ * }
+ */
+// 递归
+func defs(left *TreeNode, right *TreeNode) bool {
+ if left == nil && right == nil {
+ return true;
+ };
+ if left == nil || right == nil {
+ return false;
+ };
+ if left.Val != right.Val {
+ return false;
+ }
+ return defs(left.Left, right.Right) && defs(right.Left, left.Right);
+}
+func isSymmetric(root *TreeNode) bool {
+ return defs(root.Left, root.Right);
+}
+
+// 迭代
+func isSymmetric(root *TreeNode) bool {
+ var queue []*TreeNode;
+ if root != nil {
+ queue = append(queue, root.Left, root.Right);
+ }
+ for len(queue) > 0 {
+ left := queue[0];
+ right := queue[1];
+ queue = queue[2:];
+ if left == nil && right == nil {
+ continue;
+ }
+ if left == nil || right == nil || left.Val != right.Val {
+ return false;
+ };
+ queue = append(queue, left.Left, right.Right, right.Left, left.Right);
+ }
+ return true;
+}
+```
+
JavaScript
```javascript
From 0ddb3430986065f6a248c7936c3351544615161d Mon Sep 17 00:00:00 2001
From: "qingyi.liu"
Date: Tue, 1 Jun 2021 12:22:46 +0800
Subject: [PATCH 45/95] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=20104.=E4=BA=8C?=
=?UTF-8?q?=E5=8F=89=E6=A0=91=E7=9A=84=E6=9C=80=E5=A4=A7=E6=B7=B1=E5=BA=A6?=
=?UTF-8?q?=20go=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0104.二叉树的最大深度.md | 49 +++++++++++++++++++++++
1 file changed, 49 insertions(+)
diff --git a/problems/0104.二叉树的最大深度.md b/problems/0104.二叉树的最大深度.md
index 756afb68..5f0fe411 100644
--- a/problems/0104.二叉树的最大深度.md
+++ b/problems/0104.二叉树的最大深度.md
@@ -284,6 +284,55 @@ Python:
Go:
+```go
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ * Val int
+ * Left *TreeNode
+ * Right *TreeNode
+ * }
+ */
+func max (a, b int) int {
+ if a > b {
+ return a;
+ }
+ return b;
+}
+// 递归
+func maxDepth(root *TreeNode) int {
+ if root == nil {
+ return 0;
+ }
+ return max(maxDepth(root.Left), maxDepth(root.Right)) + 1;
+}
+
+// 遍历
+func maxDepth(root *TreeNode) int {
+ levl := 0;
+ queue := make([]*TreeNode, 0);
+ if root != nil {
+ queue = append(queue, root);
+ }
+ for l := len(queue); l > 0; {
+ for ;l > 0;l-- {
+ node := queue[0];
+ if node.Left != nil {
+ queue = append(queue, node.Left);
+ }
+ if node.Right != nil {
+ queue = append(queue, node.Right);
+ }
+ queue = queue[1:];
+ }
+ levl++;
+ l = len(queue);
+ }
+ return levl;
+}
+
+```
+
JavaScript
```javascript
From 51edb80d7c8771b0b9af932a577daeb658efd100 Mon Sep 17 00:00:00 2001
From: "qingyi.liu"
Date: Tue, 1 Jun 2021 14:20:54 +0800
Subject: [PATCH 46/95] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=20111.=E4=BA=8C?=
=?UTF-8?q?=E5=8F=89=E6=A0=91=E7=9A=84=E6=9C=80=E5=B0=8F=E6=B7=B1=E5=BA=A6?=
=?UTF-8?q?=20go=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0111.二叉树的最小深度.md | 58 +++++++++++++++++++++++
1 file changed, 58 insertions(+)
diff --git a/problems/0111.二叉树的最小深度.md b/problems/0111.二叉树的最小深度.md
index 8ee15eac..48795722 100644
--- a/problems/0111.二叉树的最小深度.md
+++ b/problems/0111.二叉树的最小深度.md
@@ -301,6 +301,64 @@ class Solution:
Go:
+```go
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ * Val int
+ * Left *TreeNode
+ * Right *TreeNode
+ * }
+ */
+func min(a, b int) int {
+ if a < b {
+ return a;
+ }
+ return b;
+}
+// 递归
+func minDepth(root *TreeNode) int {
+ if root == nil {
+ return 0;
+ }
+ if root.Left == nil && root.Right != nil {
+ return 1 + minDepth(root.Right);
+ }
+ if root.Right == nil && root.Left != nil {
+ return 1 + minDepth(root.Left);
+ }
+ return min(minDepth(root.Left), minDepth(root.Right)) + 1;
+}
+
+// 迭代
+
+func minDepth(root *TreeNode) int {
+ dep := 0;
+ queue := make([]*TreeNode, 0);
+ if root != nil {
+ queue = append(queue, root);
+ }
+ for l := len(queue); l > 0; {
+ dep++;
+ for ; l > 0; l-- {
+ node := queue[0];
+ if node.Left == nil && node.Right == nil {
+ return dep;
+ }
+ if node.Left != nil {
+ queue = append(queue, node.Left);
+ }
+ if node.Right != nil {
+ queue = append(queue, node.Right);
+ }
+ queue = queue[1:];
+ }
+ l = len(queue);
+ }
+ return dep;
+}
+```
+
JavaScript:
From ea88604bfdf249d686e1aba4a26795a35a57dcc8 Mon Sep 17 00:00:00 2001
From: Baturu <45113401+z80160280@users.noreply.github.com>
Date: Tue, 1 Jun 2021 02:44:13 -0700
Subject: [PATCH 47/95] =?UTF-8?q?Update=200459.=E9=87=8D=E5=A4=8D=E7=9A=84?=
=?UTF-8?q?=E5=AD=90=E5=AD=97=E7=AC=A6=E4=B8=B2.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0459.重复的子字符串.md | 49 ++++++++++++++++++++++++++
1 file changed, 49 insertions(+)
diff --git a/problems/0459.重复的子字符串.md b/problems/0459.重复的子字符串.md
index d9dfc62c..51a903ef 100644
--- a/problems/0459.重复的子字符串.md
+++ b/problems/0459.重复的子字符串.md
@@ -184,6 +184,55 @@ class Solution {
Python:
+这里使用了前缀表统一减一的实现方式
+
+```python
+class Solution:
+ def repeatedSubstringPattern(self, s: str) -> bool:
+ if len(s) == 0:
+ return False
+ nxt = [0] * len(s)
+ self.getNext(nxt, s)
+ if nxt[-1] != -1 and len(s) % (len(s) - (nxt[-1] + 1)) == 0:
+ return True
+ return False
+
+ def getNext(self, nxt, s):
+ nxt[0] = -1
+ j = -1
+ for i in range(1, len(s)):
+ while j >= 0 and s[i] != s[j+1]:
+ j = nxt[j]
+ if s[i] == s[j+1]:
+ j += 1
+ nxt[i] = j
+ return nxt
+```
+
+前缀表(不减一)的代码实现
+
+```python
+class Solution:
+ def repeatedSubstringPattern(self, s: str) -> bool:
+ if len(s) == 0:
+ return False
+ nxt = [0] * len(s)
+ self.getNext(nxt, s)
+ if nxt[-1] != 0 and len(s) % (len(s) - nxt[-1]) == 0:
+ return True
+ return False
+
+ def getNext(self, nxt, s):
+ nxt[0] = 0
+ j = 0
+ for i in range(1, len(s)):
+ while j > 0 and s[i] != s[j]:
+ j = nxt[j - 1]
+ if s[i] == s[j]:
+ j += 1
+ nxt[i] = j
+ return nxt
+```
Go:
From 9dd0af1946f66e1bf243d9fbc5910a9ed003a7ed Mon Sep 17 00:00:00 2001
From: youngyangyang04 <826123027@qq.com>
Date: Tue, 1 Jun 2021 17:48:26 +0800
Subject: [PATCH 48/95] Update
---
README.md | 23 ++-
problems/0001.两数之和.md | 13 +-
problems/0015.三数之和.md | 10 +-
problems/0072.编辑距离.md | 34 +++--
problems/0115.不同的子序列.md | 2 +-
problems/0121.买卖股票的最佳时机.md | 14 +-
.../0123.买卖股票的最佳时机III.md | 4 +-
problems/0416.分割等和子集.md | 7 +-
problems/0718.最长重复子数组.md | 2 +-
problems/动态规划总结篇.md | 140 ++++++++++++++++++
10 files changed, 200 insertions(+), 49 deletions(-)
create mode 100644 problems/动态规划总结篇.md
diff --git a/README.md b/README.md
index 05fc0627..b4e82eb0 100644
--- a/README.md
+++ b/README.md
@@ -112,14 +112,24 @@
(持续更新中.....)
+## 备战秋招
+
+1. [选择方向的时候,我也迷茫了](https://mp.weixin.qq.com/s/ZCzFiAHZHLqHPLJQXNm75g)
+2. [刷题就用库函数了,怎么了?](https://mp.weixin.qq.com/s/6K3_OSaudnHGq2Ey8vqYfg)
+3. [关于实习,大家可能有点迷茫!](https://mp.weixin.qq.com/s/xcxzi7c78kQGjvZ8hh7taA)
+4. [马上秋招了,慌得很!](https://mp.weixin.qq.com/s/7q7W8Cb2-a5U5atZdOnOFA)
+5. [Carl看了上百份简历,总结了这些!](https://mp.weixin.qq.com/s/sJa87MZD28piCOVMFkIbwQ)
+6. [面试中遇到了发散性问题.....](https://mp.weixin.qq.com/s/SSonDxi2pjkSVwHNzZswng)
+
## 数组
1. [数组过于简单,但你该了解这些!](./problems/数组理论基础.md)
2. [数组:每次遇到二分法,都是一看就会,一写就废](./problems/0704.二分查找.md)
3. [数组:就移除个元素很难么?](./problems/0027.移除元素.md)
-4. [数组:滑动窗口拯救了你](./problems/0209.长度最小的子数组.md)
-5. [数组:这个循环可以转懵很多人!](./problems/0059.螺旋矩阵II.md)
-6. [数组:总结篇](./problems/数组总结篇.md)
+4. [数组:有序数组的平方,还有序么?](./problems/0977.有序数组的平方.md)
+5. [数组:滑动窗口拯救了你](./problems/0209.长度最小的子数组.md)
+6. [数组:这个循环可以转懵很多人!](./problems/0059.螺旋矩阵II.md)
+7. [数组:总结篇](./problems/数组总结篇.md)
## 链表
@@ -292,6 +302,7 @@
动态规划专题已经开始啦,来不及解释了,小伙伴们上车别掉队!
+
1. [关于动态规划,你该了解这些!](./problems/动态规划理论基础.md)
2. [动态规划:斐波那契数](./problems/0509.斐波那契数.md)
3. [动态规划:爬楼梯](./problems/0070.爬楼梯.md)
@@ -366,6 +377,7 @@
52. [为了绝杀编辑距离,Carl做了三步铺垫,你都知道么?](./problems/为了绝杀编辑距离,卡尔做了三步铺垫.md)
53. [动态规划:回文子串](./problems/0647.回文子串.md)
54. [动态规划:最长回文子序列](./problems/0516.最长回文子序列.md)
+55. [动态规划总结篇](./problems/动态规划总结篇.md)
(持续更新中....)
@@ -390,12 +402,7 @@
[各类基础算法模板](https://github.com/youngyangyang04/leetcode/blob/master/problems/算法模板.md)
-# 备战秋招
-1. [技术比较弱,也对技术不感兴趣,如何选择方向?](https://mp.weixin.qq.com/s/ZCzFiAHZHLqHPLJQXNm75g)
-2. [刷题就用库函数了,怎么了?](https://mp.weixin.qq.com/s/6K3_OSaudnHGq2Ey8vqYfg)
-3. [关于实习,大家可能有点迷茫!](https://mp.weixin.qq.com/s/xcxzi7c78kQGjvZ8hh7taA)
-4. [马上秋招了,慌得很!](https://mp.weixin.qq.com/s/7q7W8Cb2-a5U5atZdOnOFA)
# B站算法视频讲解
diff --git a/problems/0001.两数之和.md b/problems/0001.两数之和.md
index 77318294..41a95daf 100644
--- a/problems/0001.两数之和.md
+++ b/problems/0001.两数之和.md
@@ -29,17 +29,17 @@ https://leetcode-cn.com/problems/two-sum/
很明显暴力的解法是两层for循环查找,时间复杂度是O(n^2)。
建议大家做这道题目之前,先做一下这两道
-* [242. 有效的字母异位词](https://mp.weixin.qq.com/s/vM6OszkM6L1Mx2Ralm9Dig)
-* [349. 两个数组的交集](https://mp.weixin.qq.com/s/N9iqAchXreSVW7zXUS4BVA)
+* [242. 有效的字母异位词](https://mp.weixin.qq.com/s/ffS8jaVFNUWyfn_8T31IdA)
+* [349. 两个数组的交集](https://mp.weixin.qq.com/s/aMSA5zrp3jJcLjuSB0Es2Q)
-[242. 有效的字母异位词](https://mp.weixin.qq.com/s/vM6OszkM6L1Mx2Ralm9Dig) 这道题目是用数组作为哈希表来解决哈希问题,[349. 两个数组的交集](https://mp.weixin.qq.com/s/N9iqAchXreSVW7zXUS4BVA)这道题目是通过set作为哈希表来解决哈希问题。
+[242. 有效的字母异位词](https://mp.weixin.qq.com/s/ffS8jaVFNUWyfn_8T31IdA) 这道题目是用数组作为哈希表来解决哈希问题,[349. 两个数组的交集](https://mp.weixin.qq.com/s/aMSA5zrp3jJcLjuSB0Es2Q)这道题目是通过set作为哈希表来解决哈希问题。
本题呢,则要使用map,那么来看一下使用数组和set来做哈希法的局限。
* 数组的大小是受限制的,而且如果元素很少,而哈希值太大会造成内存空间的浪费。
-* set是一个集合,里面放的元素只能是一个key,而两数之和这道题目,不仅要判断y是否存在而且还要记录y的下标位置,因为要返回x 和 y的下标。所以set 也不能用。
+* set是一个集合,里面放的元素只能是一个key,而两数之和这道题目,不仅要判断y是否存在而且还要记录y的下表位置,因为要返回x 和 y的下表。所以set 也不能用。
-此时就要选择另一种数据结构:map ,map是一种key value的存储结构,可以用key保存数值,用value在保存数值所在的下标。
+此时就要选择另一种数据结构:map ,map是一种key value的存储结构,可以用key保存数值,用value在保存数值所在的下表。
C++中map,有三种类型:
@@ -51,13 +51,12 @@ C++中map,有三种类型:
std::unordered_map 底层实现为哈希表,std::map 和std::multimap 的底层实现是红黑树。
-同理,std::map 和std::multimap 的key也是有序的(这个问题也经常作为面试题,考察对语言容器底层的理解)。 更多哈希表的理论知识请看[关于哈希表,你该了解这些!](https://mp.weixin.qq.com/s/g8N6WmoQmsCUw3_BaWxHZA)。
+同理,std::map 和std::multimap 的key也是有序的(这个问题也经常作为面试题,考察对语言容器底层的理解)。 更多哈希表的理论知识请看[关于哈希表,你该了解这些!](https://mp.weixin.qq.com/s/RSUANESA_tkhKhYe3ZR8Jg)。
**这道题目中并不需要key有序,选择std::unordered_map 效率更高!**
解题思路动画如下:
-

diff --git a/problems/0015.三数之和.md b/problems/0015.三数之和.md
index 4f4ec63a..5b77a170 100644
--- a/problems/0015.三数之和.md
+++ b/problems/0015.三数之和.md
@@ -105,8 +105,7 @@ public:
时间复杂度:O(n^2)。
-
-## 双指针法C++代码
+C++代码代码如下:
```C++
class Solution {
@@ -163,13 +162,14 @@ public:
# 思考题
-既然三数之和可以使用双指针法,我们之前讲过的[两数之和](https://mp.weixin.qq.com/s/uVAtjOHSeqymV8FeQbliJQ),可不可以使用双指针法呢?
+
+既然三数之和可以使用双指针法,我们之前讲过的[1.两数之和](https://mp.weixin.qq.com/s/vaMsLnH-f7_9nEK4Cuu3KQ),可不可以使用双指针法呢?
如果不能,题意如何更改就可以使用双指针法呢? **大家留言说出自己的想法吧!**
-两数之和 就不能使用双指针法,因为[两数之和](https://mp.weixin.qq.com/s/uVAtjOHSeqymV8FeQbliJQ)要求返回的是索引下表, 而双指针法一定要排序,一旦排序之后原数组的索引就被改变了。
+两数之和 就不能使用双指针法,因为[1.两数之和](https://mp.weixin.qq.com/s/vaMsLnH-f7_9nEK4Cuu3KQ)要求返回的是索引下表, 而双指针法一定要排序,一旦排序之后原数组的索引就被改变了。
-如果[两数之和](https://mp.weixin.qq.com/s/uVAtjOHSeqymV8FeQbliJQ)要求返回的是数值的话,就可以使用双指针法了。
+如果[1.两数之和](https://mp.weixin.qq.com/s/vaMsLnH-f7_9nEK4Cuu3KQ)要求返回的是数值的话,就可以使用双指针法了。
diff --git a/problems/0072.编辑距离.md b/problems/0072.编辑距离.md
index 9dd842e8..824c74af 100644
--- a/problems/0072.编辑距离.md
+++ b/problems/0072.编辑距离.md
@@ -8,6 +8,8 @@
## 72. 编辑距离
+https://leetcode-cn.com/problems/edit-distance/
+
给你两个单词 word1 和 word2,请你计算出将 word1 转换成 word2 所使用的最少操作数 。
你可以对一个单词进行如下三种操作:
@@ -16,23 +18,23 @@
* 删除一个字符
* 替换一个字符
-示例 1:
-输入:word1 = "horse", word2 = "ros"
-输出:3
-解释:
-horse -> rorse (将 'h' 替换为 'r')
-rorse -> rose (删除 'r')
-rose -> ros (删除 'e')
+示例 1:
+输入:word1 = "horse", word2 = "ros"
+输出:3
+解释:
+horse -> rorse (将 'h' 替换为 'r')
+rorse -> rose (删除 'r')
+rose -> ros (删除 'e')
-示例 2:
-输入:word1 = "intention", word2 = "execution"
-输出:5
-解释:
-intention -> inention (删除 't')
-inention -> enention (将 'i' 替换为 'e')
-enention -> exention (将 'n' 替换为 'x')
-exention -> exection (将 'n' 替换为 'c')
-exection -> execution (插入 'u')
+示例 2:
+输入:word1 = "intention", word2 = "execution"
+输出:5
+解释:
+intention -> inention (删除 't')
+inention -> enention (将 'i' 替换为 'e')
+enention -> exention (将 'n' 替换为 'x')
+exention -> exection (将 'n' 替换为 'c')
+exection -> execution (插入 'u')
提示:
diff --git a/problems/0115.不同的子序列.md b/problems/0115.不同的子序列.md
index ef098978..1661acf8 100644
--- a/problems/0115.不同的子序列.md
+++ b/problems/0115.不同的子序列.md
@@ -16,7 +16,7 @@
题目数据保证答案符合 32 位带符号整数范围。
-
+
提示:
diff --git a/problems/0121.买卖股票的最佳时机.md b/problems/0121.买卖股票的最佳时机.md
index 0544c93f..d018efb7 100644
--- a/problems/0121.买卖股票的最佳时机.md
+++ b/problems/0121.买卖股票的最佳时机.md
@@ -16,14 +16,14 @@
返回你可以从这笔交易中获取的最大利润。如果你不能获取任何利润,返回 0 。
-示例 1:
-输入:[7,1,5,3,6,4]
-输出:5
+示例 1:
+输入:[7,1,5,3,6,4]
+输出:5
解释:在第 2 天(股票价格 = 1)的时候买入,在第 5 天(股票价格 = 6)的时候卖出,最大利润 = 6-1 = 5 。注意利润不能是 7-1 = 6, 因为卖出价格需要大于买入价格;同时,你不能在买入前卖出股票。
-示例 2:
-输入:prices = [7,6,4,3,1]
-输出:0
+示例 2:
+输入:prices = [7,6,4,3,1]
+输出:0
解释:在这种情况下, 没有交易完成, 所以最大利润为 0。
@@ -33,7 +33,7 @@
这道题目最直观的想法,就是暴力,找最优间距了。
-```
+```C++
class Solution {
public:
int maxProfit(vector& prices) {
diff --git a/problems/0123.买卖股票的最佳时机III.md b/problems/0123.买卖股票的最佳时机III.md
index ecce8dc0..24370d38 100644
--- a/problems/0123.买卖股票的最佳时机III.md
+++ b/problems/0123.买卖股票的最佳时机III.md
@@ -101,9 +101,9 @@ dp[i][4] = max(dp[i - 1][4], dp[i - 1][3] + prices[i]);
所以dp[0][2] = 0;
-第0天第二次买入操作,初始值应该是多少呢?
+第0天第二次买入操作,初始值应该是多少呢?应该不少同学疑惑,第一次还没买入呢,怎么初始化第二次买入呢?
-不用管第几次,现在手头上没有现金,只要买入,现金就做相应的减少。
+第二次买入依赖于第一次卖出的状态,其实相当于第0天第一次买入了,第一次卖出了,然后在买入一次(第二次买入),那么现在手头上没有现金,只要买入,现金就做相应的减少。
所以第二次买入操作,初始化为:dp[0][3] = -prices[0];
diff --git a/problems/0416.分割等和子集.md b/problems/0416.分割等和子集.md
index 55d200e2..0d306c74 100644
--- a/problems/0416.分割等和子集.md
+++ b/problems/0416.分割等和子集.md
@@ -5,7 +5,6 @@
欢迎大家参与本项目,贡献其他语言版本的代码,拥抱开源,让更多学习算法的小伙伴们收益!
-# 动态规划:分割等和子集可以用01背包!
## 416. 分割等和子集
@@ -29,6 +28,10 @@
输出: false
解释: 数组不能分割成两个元素和相等的子集.
+提示:
+* 1 <= nums.length <= 200
+* 1 <= nums[i] <= 100
+
## 思路
这道题目初步看,是如下两题几乎是一样的,大家可以用回溯法,解决如下两题
@@ -174,7 +177,7 @@ public:
这道题目就是一道01背包应用类的题目,需要我们拆解题目,然后套入01背包的场景。
-01背包相对于本题,主要要理解,题目中物品是nums[i],重量是nums[i]i,价值也是nums[i],背包体积是sum/2。
+01背包相对于本题,主要要理解,题目中物品是nums[i],重量是nums[i],价值也是nums[i],背包体积是sum/2。
看代码的话,就可以发现,基本就是按照01背包的写法来的。
diff --git a/problems/0718.最长重复子数组.md b/problems/0718.最长重复子数组.md
index 3a377077..be9109b2 100644
--- a/problems/0718.最长重复子数组.md
+++ b/problems/0718.最长重复子数组.md
@@ -128,7 +128,7 @@ public:
**此时遍历B数组的时候,就要从后向前遍历,这样避免重复覆盖**。
-```
+```C++
class Solution {
public:
int findLength(vector& A, vector& B) {
diff --git a/problems/动态规划总结篇.md b/problems/动态规划总结篇.md
new file mode 100644
index 00000000..797f426a
--- /dev/null
+++ b/problems/动态规划总结篇.md
@@ -0,0 +1,140 @@
+
+
+
+
+
+
+
+欢迎大家参与本项目,贡献其他语言版本的代码,拥抱开源,让更多学习算法的小伙伴们收益!
+
+
+如今动态规划已经讲解了42道经典题目,共50篇文章,是时候做一篇总结了。
+
+关于动态规划,在专题第一篇[关于动态规划,你该了解这些!](https://mp.weixin.qq.com/s/ocZwfPlCWrJtVGACqFNAag)就说了动规五部曲,**而且强调了五部对解动规题目至关重要!**
+
+这是Carl做过一百多道动规题目总结出来的经验结晶啊,如果大家跟着「代码随想哦」刷过动规专题,一定会对这动规五部曲的作用感受极其深刻。
+
+动规五部曲分别为:
+
+1. 确定dp数组(dp table)以及下标的含义
+2. 确定递推公式
+3. dp数组如何初始化
+4. 确定遍历顺序
+5. 举例推导dp数组
+
+动规专题刚开始的时候,讲的题目比较简单,不少录友和我反应:这么简单的题目 讲的复杂了,不用那么多步骤分析,想出递推公式直接就AC这道题目了。
+
+**Carl的观点一直都是 简单题是用来 巩固方法论的**。 简单题目是可以靠感觉,但后面稍稍难一点的题目,估计感觉就不好使了。
+
+在动规专题讲解中,也充分体现出,这动规五部曲的重要性。
+
+还有不少录友对动规的理解是:递推公式是才是最难最重要的,只要想出递归公式,其他都好办。
+
+**其实这么想的同学基本对动规理解的不到位的**。
+
+动规五部曲里,哪一部没想清楚,这道题目基本就做不出来,即使做出来了也没有想清楚,而是朦朦胧胧的就把题目过了。
+
+* 如果想不清楚dp数组的具体含义,递归公式从何谈起,甚至初始化的时候就写错了。
+* 例如[动态规划:不同路径还不够,要有障碍!](https://mp.weixin.qq.com/s/lhqF0O4le9-wvalptOVOww) 在这道题目中,初始化才是重头戏
+* 如果看过背包系列,特别是完全背包,那么两层for循环先后顺序绝对可以搞懵很多人,反而递归公式是简单的。
+* 至于推导dp数组的重要性,动规专题里几乎每篇Carl都反复强调,当程序结果不对的时候,一定要自己推导公式,看看和程序打印的日志是否一样。
+
+好啦,我们再一起回顾一下,动态规划专题中我们都讲了哪些内容。
+
+## 动划基础
+
+* [关于动态规划,你该了解这些!](https://mp.weixin.qq.com/s/ocZwfPlCWrJtVGACqFNAag)
+* [动态规划:斐波那契数](https://mp.weixin.qq.com/s/ko0zLJplF7n_4TysnPOa_w)
+* [动态规划:爬楼梯](https://mp.weixin.qq.com/s/Ohop0jApSII9xxOMiFhGIw)
+* [动态规划:使用最小花费爬楼梯](https://mp.weixin.qq.com/s/djZB9gkyLFAKcQcSvKDorA)
+* [动态规划:不同路径](https://mp.weixin.qq.com/s/MGgGIt4QCpFMROE9X9he_A)
+* [动态规划:不同路径还不够,要有障碍!](https://mp.weixin.qq.com/s/lhqF0O4le9-wvalptOVOww)
+* [动态规划:整数拆分,你要怎么拆?](https://mp.weixin.qq.com/s/cVbyHrsWH_Rfzlj-ESr01A)
+* [动态规划:不同的二叉搜索树](https://mp.weixin.qq.com/s/8VE8pDrGxTf8NEVYBDwONw)
+
+## 背包问题系列
+
+
+
+* [动态规划:关于01背包问题,你该了解这些!](https://mp.weixin.qq.com/s/FwIiPPmR18_AJO5eiidT6w)
+* [动态规划:关于01背包问题,你该了解这些!(滚动数组)](https://mp.weixin.qq.com/s/M4uHxNVKRKm5HPjkNZBnFA)
+* [动态规划:分割等和子集可以用01背包!](https://mp.weixin.qq.com/s/sYw3QtPPQ5HMZCJcT4EaLQ)
+* [动态规划:最后一块石头的重量 II](https://mp.weixin.qq.com/s/WbwAo3jaUaNJjvhHgq0BGg)
+* [动态规划:目标和!](https://mp.weixin.qq.com/s/2pWmaohX75gwxvBENS-NCw)
+* [动态规划:一和零!](https://mp.weixin.qq.com/s/x-u3Dsp76DlYqtCe0xEKJw)
+* [动态规划:关于完全背包,你该了解这些!](https://mp.weixin.qq.com/s/akwyxlJ4TLvKcw26KB9uJw)
+* [动态规划:给你一些零钱,你要怎么凑?](https://mp.weixin.qq.com/s/PlowDsI4WMBOzf3q80AksQ)
+* [动态规划:Carl称它为排列总和!](https://mp.weixin.qq.com/s/Iixw0nahJWQgbqVNk8k6gA)
+* [动态规划:以前我没得选,现在我选择再爬一次!](https://mp.weixin.qq.com/s/e_wacnELo-2PG76EjrUakA)
+* [动态规划: 给我个机会,我再兑换一次零钱](https://mp.weixin.qq.com/s/dyk-xNilHzNtVdPPLObSeQ)
+* [动态规划:一样的套路,再求一次完全平方数](https://mp.weixin.qq.com/s/VfJT78p7UGpDZsapKF_QJQ)
+* [动态规划:单词拆分](https://mp.weixin.qq.com/s/3Spx1B6MbIYjS8YkVbByzA)
+* [动态规划:关于多重背包,你该了解这些!](https://mp.weixin.qq.com/s/b-UUUmbvG7URWyCjQkiuuQ)
+* [听说背包问题很难? 这篇总结篇来拯救你了](https://mp.weixin.qq.com/s/ZOehl3U1mDiyOQjFG1wNJA)
+
+## 打家劫舍系列
+
+* [动态规划:开始打家劫舍!](https://mp.weixin.qq.com/s/UZ31WdLEEFmBegdgLkJ8Dw)
+* [动态规划:继续打家劫舍!](https://mp.weixin.qq.com/s/kKPx4HpH3RArbRcxAVHbeQ)
+* [动态规划:还要打家劫舍!](https://mp.weixin.qq.com/s/BOJ1lHsxbQxUZffXlgglEQ)
+
+## 股票系列
+
+
+
+* [动态规划:买卖股票的最佳时机](https://mp.weixin.qq.com/s/keWo5qYJY4zmHn3amfXdfQ)
+* [动态规划:本周我们都讲了这些(系列六)](https://mp.weixin.qq.com/s/GVu-6eF0iNkpVDKRXTPOTA)
+* [动态规划:买卖股票的最佳时机II](https://mp.weixin.qq.com/s/d4TRWFuhaY83HPa6t5ZL-w)
+* [动态规划:买卖股票的最佳时机III](https://mp.weixin.qq.com/s/Sbs157mlVDtAR0gbLpdKzg)
+* [动态规划:买卖股票的最佳时机IV](https://mp.weixin.qq.com/s/jtxZJWAo2y5sUsW647Z5cw)
+* [动态规划:最佳买卖股票时机含冷冻期](https://mp.weixin.qq.com/s/TczJGFAPnkjH9ET8kwH1OA)
+* [动态规划:本周我们都讲了这些(系列七)](https://mp.weixin.qq.com/s/vdzDlrEvhXWRzblTnOnzKg)
+* [动态规划:买卖股票的最佳时机含手续费](https://mp.weixin.qq.com/s/2Cd_uINjerZ25VHH0K2IBQ)
+* [动态规划:股票系列总结篇](https://mp.weixin.qq.com/s/sC5XyEtDQWkonKnbCvZhDw)
+
+## 子序列系列
+
+
+
+* [动态规划:最长递增子序列](https://mp.weixin.qq.com/s/f8nLO3JGfgriXep_gJQpqQ)
+* [动态规划:最长连续递增序列](https://mp.weixin.qq.com/s/c0Nn0TtjkTISVdqRsyMmyA)
+* [动态规划:最长重复子数组](https://mp.weixin.qq.com/s/U5WaWqBwdoxzQDotOdWqZg)
+* [动态规划:最长公共子序列](https://mp.weixin.qq.com/s/Qq0q4HaE4TyasCTj2WGFOg)
+* [动态规划:不相交的线](https://mp.weixin.qq.com/s/krfYzSYEO8jIoVfyHzR0rw)
+* [动态规划:最大子序和](https://mp.weixin.qq.com/s/2Xtyi2L4r8sM-BcxgUKmcA)
+* [动态规划:判断子序列](https://mp.weixin.qq.com/s/2pjT4B4fjfOx5iB6N6xyng)
+* [动态规划:不同的子序列](https://mp.weixin.qq.com/s/1SULY2XVSROtk_hsoVLu8A)
+* [动态规划:两个字符串的删除操作](https://mp.weixin.qq.com/s/a8BerpqSf76DCqkPDJrpYg)
+* [动态规划:编辑距离](https://mp.weixin.qq.com/s/8aG71XjSgZG6kZbiAdkJnQ)
+* [为了绝杀编辑距离,我做了三步铺垫,你都知道么?](https://mp.weixin.qq.com/s/kbs4kCUzg8gPFttF9H3Yyw)
+* [动态规划:回文子串](https://mp.weixin.qq.com/s/2WetyP6IYQ6VotegepVpEw)
+* [动态规划:最长回文子序列](https://mp.weixin.qq.com/s/jbd3p4QPm5Kh1s2smTzWag)
+
+
+## 动规结束语
+
+关于动规,还有 树形DP(打家劫舍系列里有一道),数位DP,区间DP ,概率型DP,博弈型DP,状态压缩dp等等等,这些我就不去做讲解了,面试中出现的概率非常低。
+
+能把本篇中列举的题目都研究通透的话,你的动规水平就已经非常高了。 对付面试已经足够!
+
+这已经是全网对动规最深刻的讲解系列了。
+
+**其实大家去网上搜一搜也可以发现,能把动态规划讲清楚的资料挺少的,因为动规确实很难!要给别人讲清楚更难!**
+
+《剑指offer》上 动规的题目很少,经典的算法书籍《算法4》 没有讲 动规,而《算法导论》讲的动规基本属于劝退级别的。
+
+讲清楚一道题容易,讲清楚两道题也容易,但把整个动态规划的各个分支讲清楚,每道题目讲通透,并用一套方法论把整个动规贯彻始终就非常难了。
+
+所以Carl花费的这么大精力,把自己对动规算法理解 一五一十的全部分享给了录友们,帮助大家少走弯路!
+
+**至于动态规划PDF,即将在公众号「代码随想录」全网首发!**
+
+最后感谢录友们的一路支持,Carl才有继续更下去的动力[玫瑰],[撒花]
+
+
+-----------------------
+* 作者微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
+* B站视频:[代码随想录](https://space.bilibili.com/525438321)
+* 知识星球:[代码随想录](https://mp.weixin.qq.com/s/QVF6upVMSbgvZy8lHZS3CQ)
+
+
From 3362183d8446e551f0361ead95dca6bec8b7411a Mon Sep 17 00:00:00 2001
From: boom-jumper <56831966+boom-jumper@users.noreply.github.com>
Date: Tue, 1 Jun 2021 18:08:16 +0800
Subject: [PATCH 49/95] =?UTF-8?q?=E6=B7=BB=E5=8A=A00347=E5=89=8DK=E4=B8=AA?=
=?UTF-8?q?=E9=AB=98=E9=A2=91=E5=85=83=E7=B4=A0=20python3=20=E7=89=88?=
=?UTF-8?q?=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0347.前K个高频元素.md | 57 ++++++++++++++++++++++++++++
1 file changed, 57 insertions(+)
diff --git a/problems/0347.前K个高频元素.md b/problems/0347.前K个高频元素.md
index 841584b4..29963b48 100644
--- a/problems/0347.前K个高频元素.md
+++ b/problems/0347.前K个高频元素.md
@@ -162,6 +162,63 @@ class Solution {
Python:
+```python
+class Solution:
+ def sift(self, alist, low, high):
+ '''小根堆构建'''
+ i = low
+ j = 2 * i + 1
+ tmp = alist[low]
+ while j <= high:
+ if j + 1 <= high and alist[j+1] <= alist[j]:
+ j += 1
+ if alist[j] < tmp:
+ alist[i] = alist[j]
+ i = j
+ j = 2 * i + 1
+ else:
+ alist[i] = tmp
+ break
+ else:
+ alist[i] = tmp
+
+
+ def topK(self, nums, k):
+ # 建立小根堆
+ heap = nums[:k]
+ for i in range((k-2)//2, -1, -1):
+ self.sift(heap, i, k-1)
+
+ # 把后续的k到len(nums)填充到小根堆里
+ for i in range(k, len(nums)):
+ if nums[i] >= heap[0]:
+ heap[0] = nums[i]
+ self.sift(heap, 0, k-1)
+
+ # 排序
+ for i in range(k-1, -1, -1):
+ heap[0], heap[i]= heap[i], heap[0]
+ self.sift(heap, 0, i-1)
+ return heap
+
+ def topKFrequent(self, nums: List[int], k: int) -> List[int]:
+ dict1 = dict()
+ for val in nums:
+ if val not in dict1:
+ dict1[val] = 1
+ else:
+ dict1[val] += 1
+ res = list()
+ ind = list()
+ for item in dict1:
+ res.append([dict1[item], item])
+ result = list()
+ heap = self.topK(res, k)
+ print(heap)
+ for val in heap:
+ result.append(val[1])
+ return result
+```
Go:
From d9e4b337243d51e39bf378737c8c2eac600de3d8 Mon Sep 17 00:00:00 2001
From: Baturu <45113401+z80160280@users.noreply.github.com>
Date: Tue, 1 Jun 2021 03:47:40 -0700
Subject: [PATCH 50/95] =?UTF-8?q?Update=200239.=E6=BB=91=E5=8A=A8=E7=AA=97?=
=?UTF-8?q?=E5=8F=A3=E6=9C=80=E5=A4=A7=E5=80=BC.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0239.滑动窗口最大值.md | 35 ++++++++++++++++++++++++++
1 file changed, 35 insertions(+)
diff --git a/problems/0239.滑动窗口最大值.md b/problems/0239.滑动窗口最大值.md
index ed17157a..47383a47 100644
--- a/problems/0239.滑动窗口最大值.md
+++ b/problems/0239.滑动窗口最大值.md
@@ -263,6 +263,41 @@ class Solution {
```
Python:
+```python
+class MyQueue: #单调队列(从大到小
+ def __init__(self):
+ self.queue = [] #使用list来实现单调队列
+
+ #每次弹出的时候,比较当前要弹出的数值是否等于队列出口元素的数值,如果相等则弹出。
+ #同时pop之前判断队列当前是否为空。
+ def pop(self, value):
+ if self.queue and value == self.queue[0]:
+ self.queue.pop(0)#list.pop()时间复杂度为O(n),这里可以使用collections.deque()
+
+ #如果push的数值大于入口元素的数值,那么就将队列后端的数值弹出,直到push的数值小于等于队列入口元素的数值为止。
+ #这样就保持了队列里的数值是单调从大到小的了。
+ def push(self, value):
+ while self.queue and value > self.queue[-1]:
+ self.queue.pop()
+ self.queue.append(value)
+
+ #查询当前队列里的最大值 直接返回队列前端也就是front就可以了。
+ def front(self):
+ return self.queue[0]
+
+class Solution:
+ def maxSlidingWindow(self, nums: List[int], k: int) -> List[int]:
+ que = MyQueue()
+ result = []
+ for i in range(k): #先将前k的元素放进队列
+ que.push(nums[i])
+ result.append(que.front()) #result 记录前k的元素的最大值
+ for i in range(k, len(nums)):
+ que.pop(nums[i - k]) #滑动窗口移除最前面元素
+ que.push(nums[i]) #滑动窗口前加入最后面的元素
+ result.append(que.front()) #记录对应的最大值
+ return result
+```
Go:
From 9219242b3af749a96efbbaf7d8ea5089a36ba7fe Mon Sep 17 00:00:00 2001
From: NevS <1173325467@qq.com>
Date: Tue, 1 Jun 2021 20:34:57 +0800
Subject: [PATCH 51/95] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200001.=E4=B8=A4?=
=?UTF-8?q?=E6=95=B0=E4=B9=8B=E5=92=8C=20go=E7=89=88?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
使用map方式解题,降低时间复杂度
---
problems/0001.两数之和.md | 14 ++++++++++++++
1 file changed, 14 insertions(+)
diff --git a/problems/0001.两数之和.md b/problems/0001.两数之和.md
index 77318294..3d0674ce 100644
--- a/problems/0001.两数之和.md
+++ b/problems/0001.两数之和.md
@@ -134,6 +134,20 @@ func twoSum(nums []int, target int) []int {
}
return []int{}
}
+
+```go
+// 使用map方式解题,降低时间复杂度
+func twoSum(nums []int, target int) []int {
+ m := make(map[int]int)
+ for index, val := range nums {
+ if preIndex, ok := m[target-val]; ok {
+ return []int{preIndex, index}
+ } else {
+ m[val] = index
+ }
+ }
+ return []int{}
+}
```
Rust
From 36626386c329d75524843fab42f740e548578173 Mon Sep 17 00:00:00 2001
From: NevS <1173325467@qq.com>
Date: Tue, 1 Jun 2021 20:35:55 +0800
Subject: [PATCH 52/95] =?UTF-8?q?=E6=9B=B4=E6=94=B9=200001.=E4=B8=A4?=
=?UTF-8?q?=E6=95=B0=E4=B9=8B=E5=92=8C=20=E6=A0=BC=E5=BC=8F?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0001.两数之和.md | 1 +
1 file changed, 1 insertion(+)
diff --git a/problems/0001.两数之和.md b/problems/0001.两数之和.md
index 3d0674ce..4a1a987f 100644
--- a/problems/0001.两数之和.md
+++ b/problems/0001.两数之和.md
@@ -134,6 +134,7 @@ func twoSum(nums []int, target int) []int {
}
return []int{}
}
+```
```go
// 使用map方式解题,降低时间复杂度
From ce970df5ffdb2e428b4ec2417e2bbccdc9371a31 Mon Sep 17 00:00:00 2001
From: NevS <1173325467@qq.com>
Date: Tue, 1 Jun 2021 23:43:02 +0800
Subject: [PATCH 53/95] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200454.=E5=9B=9B?=
=?UTF-8?q?=E6=95=B0=E7=9B=B8=E5=8A=A0II=20go=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0454.四数相加II.md | 17 +++++++++++++++++
1 file changed, 17 insertions(+)
diff --git a/problems/0454.四数相加II.md b/problems/0454.四数相加II.md
index 2c648f58..5291060c 100644
--- a/problems/0454.四数相加II.md
+++ b/problems/0454.四数相加II.md
@@ -154,6 +154,23 @@ class Solution(object):
Go:
+```go
+func fourSumCount(nums1 []int, nums2 []int, nums3 []int, nums4 []int) int {
+ m := make(map[int]int)
+ count := 0
+ for _, v1 := range nums1 {
+ for _, v2 := range nums2 {
+ m[v1+v2]++
+ }
+ }
+ for _, v3 := range nums3 {
+ for _, v4 := range nums4 {
+ count += m[-v3-v4]
+ }
+ }
+ return count
+}
+```
javaScript:
From 67313906d16f8cf0e613968bf0f9b8031a273cdd Mon Sep 17 00:00:00 2001
From: Yang
Date: Tue, 1 Jun 2021 14:41:22 -0400
Subject: [PATCH 54/95] =?UTF-8?q?Update=200056.=E5=90=88=E5=B9=B6=E5=8C=BA?=
=?UTF-8?q?=E9=97=B4.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
Update Java solution. For sorting, no need to compare the second element in an interval.
---
problems/0056.合并区间.md | 6 +-----
1 file changed, 1 insertion(+), 5 deletions(-)
diff --git a/problems/0056.合并区间.md b/problems/0056.合并区间.md
index eb8e7bff..f42fced7 100644
--- a/problems/0056.合并区间.md
+++ b/problems/0056.合并区间.md
@@ -144,11 +144,7 @@ class Solution {
Arrays.sort(intervals, new Comparator() {
@Override
public int compare(int[] o1, int[] o2) {
- if (o1[0] != o2[0]) {
- return Integer.compare(o1[0],o2[0]);
- } else {
- return Integer.compare(o1[1],o2[1]);
- }
+ return Integer.compare(o1[0], o2[0]);
}
});
From 838e290312af9218c62630a18022087205c8a2bc Mon Sep 17 00:00:00 2001
From: Yang
Date: Tue, 1 Jun 2021 14:46:19 -0400
Subject: [PATCH 55/95] =?UTF-8?q?Update=200056.=E5=90=88=E5=B9=B6=E5=8C=BA?=
=?UTF-8?q?=E9=97=B4.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
Simplify sorting
---
problems/0056.合并区间.md | 7 +------
1 file changed, 1 insertion(+), 6 deletions(-)
diff --git a/problems/0056.合并区间.md b/problems/0056.合并区间.md
index f42fced7..43ff447d 100644
--- a/problems/0056.合并区间.md
+++ b/problems/0056.合并区间.md
@@ -141,12 +141,7 @@ Java:
class Solution {
public int[][] merge(int[][] intervals) {
List res = new LinkedList<>();
- Arrays.sort(intervals, new Comparator() {
- @Override
- public int compare(int[] o1, int[] o2) {
- return Integer.compare(o1[0], o2[0]);
- }
- });
+ Arrays.sort(intervals, (o1, o2) -> o1[0] - o2[0]);
int start = intervals[0][0];
for (int i = 1; i < intervals.length; i++) {
From f4736c7dd2fb5ce75864dc34baf5e4bf212c16d4 Mon Sep 17 00:00:00 2001
From: Yang
Date: Tue, 1 Jun 2021 14:48:52 -0400
Subject: [PATCH 56/95] Improve comparison
---
problems/0056.合并区间.md | 2 +-
1 file changed, 1 insertion(+), 1 deletion(-)
diff --git a/problems/0056.合并区间.md b/problems/0056.合并区间.md
index 43ff447d..d4ffc554 100644
--- a/problems/0056.合并区间.md
+++ b/problems/0056.合并区间.md
@@ -141,7 +141,7 @@ Java:
class Solution {
public int[][] merge(int[][] intervals) {
List res = new LinkedList<>();
- Arrays.sort(intervals, (o1, o2) -> o1[0] - o2[0]);
+ Arrays.sort(intervals, (o1, o2) -> Integer.compare(o1[0], o2[0]));
int start = intervals[0][0];
for (int i = 1; i < intervals.length; i++) {
From bb5cfa68b2a20bc3609b56cd65d2542630c86f37 Mon Sep 17 00:00:00 2001
From: "Neil.Liu" <88214924@qq.com>
Date: Wed, 2 Jun 2021 10:15:47 +0800
Subject: [PATCH 57/95] =?UTF-8?q?=E6=B7=BB=E5=8A=A00151.=E7=BF=BB=E8=BD=AC?=
=?UTF-8?q?=E5=AD=97=E7=AC=A6=E4=B8=B2=E9=87=8C=E7=9A=84=E5=8D=95=E8=AF=8D?=
=?UTF-8?q?Go=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0151.翻转字符串里的单词.md | 56 +++++++++++++++++++-
1 file changed, 54 insertions(+), 2 deletions(-)
diff --git a/problems/0151.翻转字符串里的单词.md b/problems/0151.翻转字符串里的单词.md
index 512360fe..63499b71 100644
--- a/problems/0151.翻转字符串里的单词.md
+++ b/problems/0151.翻转字符串里的单词.md
@@ -318,9 +318,61 @@ class Solution {
Python:
-
Go:
+```go
+import (
+ "fmt"
+)
+
+func reverseWords(s string) string {
+ //1.使用双指针删除冗余的空格
+ slowIndex, fastIndex := 0, 0
+ b := []byte(s)
+ //删除头部冗余空格
+ for len(b) > 0 && fastIndex < len(b) && b[fastIndex] == ' ' {
+ fastIndex++
+ }
+ //删除单词间冗余空格
+ for ; fastIndex < len(b); fastIndex++ {
+ if fastIndex-1 > 0 && b[fastIndex-1] == b[fastIndex] && b[fastIndex] == ' ' {
+ continue
+ }
+ b[slowIndex] = b[fastIndex]
+ slowIndex++
+ }
+ //删除尾部冗余空格
+ if slowIndex-1 > 0 && b[slowIndex-1] == ' ' {
+ b = b[:slowIndex-1]
+ } else {
+ b = b[:slowIndex]
+ }
+ //2.反转整个字符串
+ reverse(&b, 0, len(b)-1)
+ //3.反转单个单词 i单词开始位置,j单词结束位置
+ i := 0
+ for i < len(b) {
+ j := i
+ for ; j < len(b) && b[j] != ' '; j++ {
+ }
+ reverse(&b, i, j-1)
+ i = j
+ i++
+ }
+ return string(b)
+}
+
+func reverse(b *[]byte, left, right int) {
+ for left < right {
+ (*b)[left], (*b)[right] = (*b)[right], (*b)[left]
+ left++
+ right--
+ }
+}
+```
+
+
+
@@ -328,4 +380,4 @@ Go:
* 作者微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
* B站视频:[代码随想录](https://space.bilibili.com/525438321)
* 知识星球:[代码随想录](https://mp.weixin.qq.com/s/QVF6upVMSbgvZy8lHZS3CQ)
-
+
\ No newline at end of file
From fae5373477b50d259c94c08dcef1ce6254081491 Mon Sep 17 00:00:00 2001
From: "Neil.Liu" <88214924@qq.com>
Date: Wed, 2 Jun 2021 10:25:09 +0800
Subject: [PATCH 58/95] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=E5=89=91=E6=8C=87Offer?=
=?UTF-8?q?58-II.=E5=B7=A6=E6=97=8B=E8=BD=AC=E5=AD=97=E7=AC=A6=E4=B8=B2Go?=
=?UTF-8?q?=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
.../剑指Offer58-II.左旋转字符串.md | 28 +++++++++++++++++--
1 file changed, 25 insertions(+), 3 deletions(-)
diff --git a/problems/剑指Offer58-II.左旋转字符串.md b/problems/剑指Offer58-II.左旋转字符串.md
index 648546c1..39c8382c 100644
--- a/problems/剑指Offer58-II.左旋转字符串.md
+++ b/problems/剑指Offer58-II.左旋转字符串.md
@@ -23,7 +23,7 @@ https://leetcode-cn.com/problems/zuo-xuan-zhuan-zi-fu-chuan-lcof/
示例 2:
输入: s = "lrloseumgh", k = 6
输出: "umghlrlose"
-
+
限制:
1 <= k < s.length <= 10000
@@ -119,9 +119,31 @@ class Solution {
```
Python:
-
Go:
+```go
+func reverseLeftWords(s string, n int) string {
+ b := []byte(s)
+ // 1. 反转前n个字符
+ // 2. 反转第n到end字符
+ // 3. 反转整个字符
+ reverse(b, 0, n-1)
+ reverse(b, n, len(b)-1)
+ reverse(b, 0, len(b)-1)
+ return string(b)
+}
+// 切片是引用传递
+func reverse(b []byte, left, right int){
+ for left < right{
+ b[left], b[right] = b[right],b[left]
+ left++
+ right--
+ }
+}
+```
+
+
+
@@ -129,4 +151,4 @@ Go:
* 作者微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
* B站视频:[代码随想录](https://space.bilibili.com/525438321)
* 知识星球:[代码随想录](https://mp.weixin.qq.com/s/QVF6upVMSbgvZy8lHZS3CQ)
-
+
\ No newline at end of file
From 9163ed6b65667e4c95a9ed2b26bd63a6a516fc9f Mon Sep 17 00:00:00 2001
From: zhangzw
Date: Wed, 2 Jun 2021 11:49:59 +0800
Subject: [PATCH 59/95] =?UTF-8?q?=E6=B7=BB=E5=8A=A01049.=E6=9C=80=E5=90=8E?=
=?UTF-8?q?=E4=B8=80=E5=9D=97=E7=9F=B3=E5=A4=B4=E7=9A=84=E9=87=8D=E9=87=8F?=
=?UTF-8?q?II=20Go=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
.../1049.最后一块石头的重量II.md | 26 +++++++++++++++++++
1 file changed, 26 insertions(+)
diff --git a/problems/1049.最后一块石头的重量II.md b/problems/1049.最后一块石头的重量II.md
index f74600b1..c09e476a 100644
--- a/problems/1049.最后一块石头的重量II.md
+++ b/problems/1049.最后一块石头的重量II.md
@@ -191,7 +191,33 @@ class Solution:
```
Go:
+```go
+func lastStoneWeightII(stones []int) int {
+ // 15001 = 30 * 1000 /2 +1
+ dp := make([]int, 15001)
+ // 求target
+ sum := 0
+ for _, v := range stones {
+ sum += v
+ }
+ target := sum / 2
+ // 遍历顺序
+ for i := 0; i < len(stones); i++ {
+ for j := target; j >= stones[i]; j-- {
+ // 推导公式
+ dp[j] = max(dp[j], dp[j-stones[i]]+stones[i])
+ }
+ }
+ return sum - 2 * dp[target]
+}
+func max(a, b int) int {
+ if a > b {
+ return a
+ }
+ return b
+}
+```
From 9f84671e690208a1b40debe15b1bfa624448f348 Mon Sep 17 00:00:00 2001
From: zhangzw
Date: Wed, 2 Jun 2021 11:51:50 +0800
Subject: [PATCH 60/95] =?UTF-8?q?=E6=B7=BB=E5=8A=A00494.=E7=9B=AE=E6=A0=87?=
=?UTF-8?q?=E5=92=8C=20Go=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0494.目标和.md | 30 +++++++++++++++++++++++++++++-
1 file changed, 29 insertions(+), 1 deletion(-)
diff --git a/problems/0494.目标和.md b/problems/0494.目标和.md
index c772a09d..c917ed5c 100644
--- a/problems/0494.目标和.md
+++ b/problems/0494.目标和.md
@@ -276,7 +276,35 @@ class Solution:
```
Go:
-
+```go
+func findTargetSumWays(nums []int, target int) int {
+ sum := 0
+ for _, v := range nums {
+ sum += v
+ }
+ if target > sum {
+ return 0
+ }
+ if (sum+target)%2 == 1 {
+ return 0
+ }
+ // 计算背包大小
+ bag := (sum + target) / 2
+ // 定义dp数组
+ dp := make([]int, bag+1)
+ // 初始化
+ dp[0] = 1
+ // 遍历顺序
+ for i := 0; i < len(nums); i++ {
+ for j := bag; j >= nums[i]; j-- {
+ //推导公式
+ dp[j] += dp[j-nums[i]]
+ //fmt.Println(dp)
+ }
+ }
+ return dp[bag]
+}
+```
From 9707fbc521b16e79a43ef588e4f4a77454a30fdf Mon Sep 17 00:00:00 2001
From: Baturu <45113401+z80160280@users.noreply.github.com>
Date: Wed, 2 Jun 2021 02:41:59 -0700
Subject: [PATCH 61/95] =?UTF-8?q?Update=200347.=E5=89=8DK=E4=B8=AA?=
=?UTF-8?q?=E9=AB=98=E9=A2=91=E5=85=83=E7=B4=A0.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0347.前K个高频元素.md | 28 +++++++++++++++++++++++++++-
1 file changed, 27 insertions(+), 1 deletion(-)
diff --git a/problems/0347.前K个高频元素.md b/problems/0347.前K个高频元素.md
index 841584b4..71af618e 100644
--- a/problems/0347.前K个高频元素.md
+++ b/problems/0347.前K个高频元素.md
@@ -162,7 +162,33 @@ class Solution {
Python:
-
+```python
+#时间复杂度:O(nlogk)
+#空间复杂度:O(n)
+import heapq
+class Solution:
+ def topKFrequent(self, nums: List[int], k: int) -> List[int]:
+ #要统计元素出现频率
+ map_ = {} #nums[i]:对应出现的次数
+ for i in range(len(nums)):
+ map_[nums[i]] = map_.get(nums[i], 0) + 1
+
+ #对频率排序
+ #定义一个小顶堆,大小为k
+ pri_que = [] #小顶堆
+
+ #用固定大小为k的小顶堆,扫面所有频率的数值
+ for key, freq in map_.items():
+ heapq.heappush(pri_que, (freq, key))
+ if len(pri_que) > k: #如果堆的大小大于了K,则队列弹出,保证堆的大小一直为k
+ heapq.heappop(pri_que)
+
+ #找出前K个高频元素,因为小顶堆先弹出的是最小的,所以倒叙来输出到数组
+ result = [0] * k
+ for i in range(k-1, -1, -1):
+ result[i] = heapq.heappop(pri_que)[1]
+ return result
+```
Go:
From 2a7c3cb87c2fbcefe1b1995cfa0d45ba7e857627 Mon Sep 17 00:00:00 2001
From: Baturu <45113401+z80160280@users.noreply.github.com>
Date: Wed, 2 Jun 2021 03:09:05 -0700
Subject: [PATCH 62/95] =?UTF-8?q?Update=20=E4=BA=8C=E5=8F=89=E6=A0=91?=
=?UTF-8?q?=E7=9A=84=E7=BB=9F=E4=B8=80=E8=BF=AD=E4=BB=A3=E6=B3=95.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/二叉树的统一迭代法.md | 71 +++++++++++++++++++++++++
1 file changed, 71 insertions(+)
diff --git a/problems/二叉树的统一迭代法.md b/problems/二叉树的统一迭代法.md
index dca5d3e3..dc79643d 100644
--- a/problems/二叉树的统一迭代法.md
+++ b/problems/二叉树的统一迭代法.md
@@ -239,7 +239,78 @@ Java:
```
Python:
+> 迭代法前序遍历
+```python
+class Solution:
+ def preorderTraversal(self, root: TreeNode) -> List[int]:
+ result = []
+ st= []
+ if root:
+ st.append(root)
+ while st:
+ node = st.pop()
+ if node != None:
+ if node.right: #右
+ st.append(node.right)
+ if node.left: #左
+ st.append(node.left)
+ st.append(node) #中
+ st.append(None)
+ else:
+ node = st.pop()
+ result.append(node.val)
+ return result
+```
+
+> 迭代法中序遍历
+```python
+class Solution:
+ def inorderTraversal(self, root: TreeNode) -> List[int]:
+ result = []
+ st = []
+ if root:
+ st.append(root)
+ while st:
+ node = st.pop()
+ if node != None:
+ if node.right: #添加右节点(空节点不入栈)
+ st.append(node.right)
+
+ st.append(node) #添加中节点
+ st.append(None) #中节点访问过,但是还没有处理,加入空节点做为标记。
+
+ if node.left: #添加左节点(空节点不入栈)
+ st.append(node.left)
+ else: #只有遇到空节点的时候,才将下一个节点放进结果集
+ node = st.pop() #重新取出栈中元素
+ result.append(node.val) #加入到结果集
+ return result
+```
+
+> 迭代法后序遍历
+```python
+class Solution:
+ def postorderTraversal(self, root: TreeNode) -> List[int]:
+ result = []
+ st = []
+ if root:
+ st.append(root)
+ while st:
+ node = st.pop()
+ if node != None:
+ st.append(node) #中
+ st.append(None)
+
+ if node.right: #右
+ st.append(node.right)
+ if node.left: #左
+ st.append(node.left)
+ else:
+ node = st.pop()
+ result.append(node.val)
+ return result
+```
Go:
> 前序遍历统一迭代法
From 045f3f98d6234e548c8b98edee911f2eb39b482f Mon Sep 17 00:00:00 2001
From: fusunx <1102654482@qq.com>
Date: Wed, 2 Jun 2021 20:10:28 +0800
Subject: [PATCH 63/95] =?UTF-8?q?0134.=E5=8A=A0=E6=B2=B9=E7=AB=99.md=20Jav?=
=?UTF-8?q?ascript?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0134.加油站.md | 21 +++++++++++++++++++++
1 file changed, 21 insertions(+)
diff --git a/problems/0134.加油站.md b/problems/0134.加油站.md
index 5c3f70a8..dfed2d96 100644
--- a/problems/0134.加油站.md
+++ b/problems/0134.加油站.md
@@ -241,7 +241,28 @@ class Solution:
Go:
+Javascript:
+```Javascript
+var canCompleteCircuit = function(gas, cost) {
+ const gasLen = gas.length
+ let start = 0
+ let curSum = 0
+ let totalSum = 0
+ for(let i = 0; i < gasLen; i++) {
+ curSum += gas[i] - cost[i]
+ totalSum += gas[i] - cost[i]
+ if(curSum < 0) {
+ curSum = 0
+ start = i + 1
+ }
+ }
+
+ if(totalSum < 0) return -1
+
+ return start
+};
+```
-----------------------
From bed3f0b09fc46f11d319f33225e78e7c12d5842b Mon Sep 17 00:00:00 2001
From: NevS <1173325467@qq.com>
Date: Wed, 2 Jun 2021 20:11:11 +0800
Subject: [PATCH 64/95] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200383.=E8=B5=8E?=
=?UTF-8?q?=E9=87=91=E4=BF=A1=20go=E7=89=88?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0383.赎金信.md | 15 +++++++++++++++
1 file changed, 15 insertions(+)
diff --git a/problems/0383.赎金信.md b/problems/0383.赎金信.md
index fd8175db..23e2c5fd 100644
--- a/problems/0383.赎金信.md
+++ b/problems/0383.赎金信.md
@@ -166,6 +166,21 @@ class Solution(object):
```
Go:
+```go
+func canConstruct(ransomNote string, magazine string) bool {
+ record := make([]int, 26)
+ for _, v := range magazine {
+ record[v-'a']++
+ }
+ for _, v := range ransomNote {
+ record[v-'a']--
+ if record[v-'a'] < 0 {
+ return false
+ }
+ }
+ return true
+}
+```
javaScript:
From 41c4f17833df822a740ae2b901fb3bb530b5b47b Mon Sep 17 00:00:00 2001
From: evanlai
Date: Wed, 2 Jun 2021 23:39:59 +0800
Subject: [PATCH 65/95] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200455.=E5=88=86?=
=?UTF-8?q?=E5=8F=91=E9=A5=BC=E5=B9=B2=20python3=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0455.分发饼干.md | 16 +++++++++++++++-
1 file changed, 15 insertions(+), 1 deletion(-)
diff --git a/problems/0455.分发饼干.md b/problems/0455.分发饼干.md
index b3714c43..ecf7f132 100644
--- a/problems/0455.分发饼干.md
+++ b/problems/0455.分发饼干.md
@@ -134,7 +134,21 @@ class Solution {
```
Python:
-
+```python
+class Solution:
+ def findContentChildren(self, g: List[int], s: List[int]) -> int:
+ #先考虑胃口小的孩子
+ g.sort()
+ s.sort()
+ i=j=0
+ count = 0
+ while(i
Date: Wed, 2 Jun 2021 23:41:29 +0800
Subject: [PATCH 66/95] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200376.=E6=91=86?=
=?UTF-8?q?=E5=8A=A8=E5=BA=8F=E5=88=97=20python3=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0376.摆动序列.md | 16 +++++++++++++++-
1 file changed, 15 insertions(+), 1 deletion(-)
diff --git a/problems/0376.摆动序列.md b/problems/0376.摆动序列.md
index 6fa30cde..bbca5ea0 100644
--- a/problems/0376.摆动序列.md
+++ b/problems/0376.摆动序列.md
@@ -138,7 +138,21 @@ class Solution {
```
Python:
-
+```python
+class Solution:
+ def wiggleMaxLength(self, nums: List[int]) -> int:
+ #贪心 求波峰数量 + 波谷数量
+ if len(nums)<=1:
+ return len(nums)
+ cur, pre = 0,0 #当前一对差值,前一对差值
+ count = 1#默认最右边有一个峰值
+ for i in range(len(nums)-1):
+ cur = nums[i+1] - nums[i]
+ if((cur>0 and pre<=0) or (cur<0 and pre>=0)):
+ count += 1
+ pre = cur
+ return count
+```
Go:
From 37515f2202b8a9868a25f8c000f46fe589a093a3 Mon Sep 17 00:00:00 2001
From: fusunx <1102654482@qq.com>
Date: Thu, 3 Jun 2021 08:25:39 +0800
Subject: [PATCH 67/95] =?UTF-8?q?0135.=E5=88=86=E5=8F=91=E7=B3=96=E6=9E=9C?=
=?UTF-8?q?.md=20Javascript?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0135.分发糖果.md | 23 +++++++++++++++++++++++
1 file changed, 23 insertions(+)
diff --git a/problems/0135.分发糖果.md b/problems/0135.分发糖果.md
index a3311791..fd791277 100644
--- a/problems/0135.分发糖果.md
+++ b/problems/0135.分发糖果.md
@@ -176,7 +176,30 @@ class Solution:
Go:
+Javascript:
+```Javascript
+var candy = function(ratings) {
+ let candys = new Array(ratings.length).fill(1)
+ for(let i = 1; i < ratings.length; i++) {
+ if(ratings[i] > ratings[i - 1]) {
+ candys[i] = candys[i - 1] + 1
+ }
+ }
+
+ for(let i = ratings.length - 2; i >= 0; i--) {
+ if(ratings[i] > ratings[i + 1]) {
+ candys[i] = Math.max(candys[i], candys[i + 1] + 1)
+ }
+ }
+
+ let count = candys.reduce((a, b) => {
+ return a + b
+ })
+
+ return count
+};
+```
-----------------------
From 1a4a287243e43649166fea0d028dc8b733602387 Mon Sep 17 00:00:00 2001
From: nmydt <62681228+nmydt@users.noreply.github.com>
Date: Thu, 3 Jun 2021 13:00:14 +0800
Subject: [PATCH 68/95] =?UTF-8?q?=E5=A2=9E=E5=8A=A0112.=E8=B7=AF=E5=BE=84?=
=?UTF-8?q?=E6=80=BB=E5=92=8C=20JAVA=20=E8=BF=AD=E4=BB=A3=E7=89=88?=
=?UTF-8?q?=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0112.路径总和.md | 30 ++++++++++++++++++++++++++++++
1 file changed, 30 insertions(+)
diff --git a/problems/0112.路径总和.md b/problems/0112.路径总和.md
index 4ccd8912..d810a046 100644
--- a/problems/0112.路径总和.md
+++ b/problems/0112.路径总和.md
@@ -345,6 +345,36 @@ class Solution {
return hasPathSum(root.left, targetSum - root.val) || hasPathSum(root.right, targetSum - root.val);
}
}
+```
+迭代
+```java
+class Solution {
+ public boolean hasPathSum(TreeNode root, int targetSum) {
+ if(root==null)return false;
+ Stack stack1 = new Stack<>();
+ Stack stack2 = new Stack<>();
+ stack1.push(root);stack2.push(root.val);
+ while(!stack1.isEmpty()){
+ int size = stack1.size();
+ for(int i=0;i
Date: Thu, 3 Jun 2021 14:30:31 +0800
Subject: [PATCH 69/95] =?UTF-8?q?=E6=B7=BB=E5=8A=A00028.=E5=AE=9E=E7=8E=B0?=
=?UTF-8?q?strStr=20Go=E8=AF=AD=E8=A8=80=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0028.实现strStr.md | 90 ++++++++++++++++++++++++++++++++++-
1 file changed, 89 insertions(+), 1 deletion(-)
diff --git a/problems/0028.实现strStr.md b/problems/0028.实现strStr.md
index b8ebcaa1..1dba5a38 100644
--- a/problems/0028.实现strStr.md
+++ b/problems/0028.实现strStr.md
@@ -726,10 +726,98 @@ class Solution:
Go:
+```go
+// 方法一:前缀表使用减1实现
+
+// getNext 构造前缀表next
+// params:
+// next 前缀表数组
+// s 模式串
+func getNext(next []int, s string) {
+ j := -1 // j表示 最长相等前后缀长度
+ next[0] = j
+
+ for i := 1; i < len(s); i++ {
+ for j >= 0 && s[i] != s[j+1] {
+ j = next[j] // 回退前一位
+ }
+ if s[i] == s[j+1] {
+ j++
+ }
+ next[i] = j // next[i]是i(包括i)之前的最长相等前后缀长度
+ }
+}
+func strStr(haystack string, needle string) int {
+ if len(needle) == 0 {
+ return 0
+ }
+ next := make([]int, len(needle))
+ getNext(next, needle)
+ j := -1 // 模式串的起始位置 next为-1 因此也为-1
+ for i := 0; i < len(haystack); i++ {
+ for j >= 0 && haystack[i] != needle[j+1] {
+ j = next[j] // 寻找下一个匹配点
+ }
+ if haystack[i] == needle[j+1] {
+ j++
+ }
+ if j == len(needle)-1 { // j指向了模式串的末尾
+ return i - len(needle) + 1
+ }
+ }
+ return -1
+}
+```
+
+```go
+// 方法二: 前缀表无减一或者右移
+
+// getNext 构造前缀表next
+// params:
+// next 前缀表数组
+// s 模式串
+func getNext(next []int, s string) {
+ j := 0
+ next[0] = j
+ for i := 1; i < len(s); i++ {
+ for j > 0 && s[i] != s[j] {
+ j = next[j-1]
+ }
+ if s[i] == s[j] {
+ j++
+ }
+ next[i] = j
+ }
+}
+func strStr(haystack string, needle string) int {
+ n := len(needle)
+ if n == 0 {
+ return 0
+ }
+ j := 0
+ next := make([]int, n)
+ getNext(next, needle)
+ for i := 0; i < len(haystack); i++ {
+ for j > 0 && haystack[i] != needle[j] {
+ j = next[j-1] // 回退到j的前一位
+ }
+ if haystack[i] == needle[j] {
+ j++
+ }
+ if j == n {
+ return i - n + 1
+ }
+ }
+ return -1
+}
+```
+
+
+
-----------------------
* 作者微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
* B站视频:[代码随想录](https://space.bilibili.com/525438321)
* 知识星球:[代码随想录](https://mp.weixin.qq.com/s/QVF6upVMSbgvZy8lHZS3CQ)
-
+
\ No newline at end of file
From 7bf42a83f42131f87c881a42567994379ceb6b8c Mon Sep 17 00:00:00 2001
From: "Neil.Liu" <88214924@qq.com>
Date: Thu, 3 Jun 2021 15:42:44 +0800
Subject: [PATCH 70/95] =?UTF-8?q?=E6=B7=BB=E5=8A=A00459.=E9=87=8D=E5=A4=8D?=
=?UTF-8?q?=E7=9A=84=E5=AD=90=E5=AD=97=E7=AC=A6=E4=B8=B2=20Go=E8=AF=AD?=
=?UTF-8?q?=E8=A8=80=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0459.重复的子字符串.md | 60 +++++++++++++++++++++++++-
1 file changed, 59 insertions(+), 1 deletion(-)
diff --git a/problems/0459.重复的子字符串.md b/problems/0459.重复的子字符串.md
index 51a903ef..deb755bf 100644
--- a/problems/0459.重复的子字符串.md
+++ b/problems/0459.重复的子字符串.md
@@ -236,6 +236,64 @@ class Solution:
Go:
+这里使用了前缀表统一减一的实现方式
+
+```go
+func repeatedSubstringPattern(s string) bool {
+ n := len(s)
+ if n == 0 {
+ return false
+ }
+ next := make([]int, n)
+ j := -1
+ next[0] = j
+ for i := 1; i < n; i++ {
+ for j >= 0 && s[i] != s[j+1] {
+ j = next[j]
+ }
+ if s[i] == s[j+1] {
+ j++
+ }
+ next[i] = j
+ }
+ // next[n-1]+1 最长相同前后缀的长度
+ if next[n-1] != -1 && n%(n-(next[n-1]+1)) == 0 {
+ return true
+ }
+ return false
+}
+```
+
+前缀表(不减一)的代码实现
+
+```go
+func repeatedSubstringPattern(s string) bool {
+ n := len(s)
+ if n == 0 {
+ return false
+ }
+ j := 0
+ next := make([]int, n)
+ next[0] = j
+ for i := 1; i < n; i++ {
+ for j > 0 && s[i] != s[j] {
+ j = next[j-1]
+ }
+ if s[i] == s[j] {
+ j++
+ }
+ next[i] = j
+ }
+ // next[n-1] 最长相同前后缀的长度
+ if next[n-1] != 0 && n%(n-next[n-1]) == 0 {
+ return true
+ }
+ return false
+}
+```
+
+
+
@@ -243,4 +301,4 @@ Go:
* 作者微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
* B站视频:[代码随想录](https://space.bilibili.com/525438321)
* 知识星球:[代码随想录](https://mp.weixin.qq.com/s/QVF6upVMSbgvZy8lHZS3CQ)
-
+
\ No newline at end of file
From 36957838cb9b449ba118d5762e401933258acdf8 Mon Sep 17 00:00:00 2001
From: jojoo15 <75017412+jojoo15@users.noreply.github.com>
Date: Thu, 3 Jun 2021 10:41:52 +0200
Subject: [PATCH 71/95] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200037.=E8=A7=A3?=
=?UTF-8?q?=E6=95=B0=E7=8B=AC=20python3=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
添加 0037.解数独 python3版本
---
problems/0037.解数独.md | 34 +++++++++++++++++++++++++++++++++-
1 file changed, 33 insertions(+), 1 deletion(-)
diff --git a/problems/0037.解数独.md b/problems/0037.解数独.md
index d5df95e8..4eb60704 100644
--- a/problems/0037.解数独.md
+++ b/problems/0037.解数独.md
@@ -287,7 +287,39 @@ class Solution {
```
Python:
-
+```python3
+class Solution:
+ def solveSudoku(self, board: List[List[str]]) -> None:
+ """
+ Do not return anything, modify board in-place instead.
+ """
+ def backtrack(board):
+ for i in range(len(board)): #遍历行
+ for j in range(len(board[0])): #遍历列
+ if board[i][j] != ".": continue
+ for k in range(1,10): #(i, j) 这个位置放k是否合适
+ if isValid(i,j,k,board):
+ board[i][j] = str(k) #放置k
+ if backtrack(board): return True #如果找到合适一组立刻返回
+ board[i][j] = "." #回溯,撤销k
+ return False #9个数都试完了,都不行,那么就返回false
+ return True #遍历完没有返回false,说明找到了合适棋盘位置了
+ def isValid(row,col,val,board):
+ for i in range(9): #判断行里是否重复
+ if board[row][i] == str(val):
+ return False
+ for j in range(9): #判断列里是否重复
+ if board[j][col] == str(val):
+ return False
+ startRow = (row // 3) * 3
+ startcol = (col // 3) * 3
+ for i in range(startRow,startRow + 3): #判断9方格里是否重复
+ for j in range(startcol,startcol + 3):
+ if board[i][j] == str(val):
+ return False
+ return True
+ backtrack(board)
+```
Go:
From 17dd8f8499e6f5ab65d0529b3b9d9b388479b680 Mon Sep 17 00:00:00 2001
From: NevS <1173325467@qq.com>
Date: Fri, 4 Jun 2021 00:20:25 +0800
Subject: [PATCH 72/95] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200541.=E5=8F=8D?=
=?UTF-8?q?=E8=BD=AC=E5=AD=97=E7=AC=A6=E4=B8=B2II=20go=E7=89=88?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0541.反转字符串II.md | 24 ++++++++++++++++++++++++
1 file changed, 24 insertions(+)
diff --git a/problems/0541.反转字符串II.md b/problems/0541.反转字符串II.md
index 6c8f3e94..00581fc0 100644
--- a/problems/0541.反转字符串II.md
+++ b/problems/0541.反转字符串II.md
@@ -164,6 +164,30 @@ class Solution(object):
Go:
+```go
+func reverseStr(s string, k int) string {
+ ss := []byte(s)
+ length := len(s)
+ for i := 0; i < length; i += 2 * k {
+ if i + k <= length {
+ reverse(ss[i:i+k])
+ } else {
+ reverse(ss[i:length])
+ }
+ }
+ return string(ss)
+}
+
+func reverse(b []byte) {
+ left := 0
+ right := len(b) - 1
+ for left < right {
+ b[left], b[right] = b[right], b[left]
+ left++
+ right--
+ }
+}
+```
javaScript:
From 0dd4590945249059a37638bfe3ab79dc134b8e40 Mon Sep 17 00:00:00 2001
From: Yang Li
Date: Thu, 3 Jun 2021 18:44:16 -0400
Subject: [PATCH 73/95] =?UTF-8?q?Update=20=E9=93=BE=E8=A1=A8=E7=90=86?=
=?UTF-8?q?=E8=AE=BA=E5=9F=BA=E7=A1=80.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
fix typo
---
problems/链表理论基础.md | 2 +-
1 file changed, 1 insertion(+), 1 deletion(-)
diff --git a/problems/链表理论基础.md b/problems/链表理论基础.md
index ea3b9098..252247c7 100644
--- a/problems/链表理论基础.md
+++ b/problems/链表理论基础.md
@@ -83,7 +83,7 @@ struct ListNode {
有同学说了,我不定义构造函数行不行,答案是可以的,C++默认生成一个构造函数。
-但是这个构造函数不会初始化任何成员变化,下面我来举两个例子:
+但是这个构造函数不会初始化任何成员变量,下面我来举两个例子:
通过自己定义构造函数初始化节点:
From b503c622640ca966325a37b96b63080a3bac1972 Mon Sep 17 00:00:00 2001
From: youngyangyang04 <826123027@qq.com>
Date: Fri, 4 Jun 2021 09:30:11 +0800
Subject: [PATCH 74/95] Update
---
README.md | 6 +-
problems/0018.四数之和.md | 27 +-
problems/0028.实现strStr.md | 5 -
problems/0454.四数相加II.md | 3 +-
problems/二叉树的理论基础.md | 202 -----------
.../背包理论基础01背包-一维DP.md | 231 -------------
.../背包理论基础01背包-二维DP.md | 318 ------------------
7 files changed, 18 insertions(+), 774 deletions(-)
delete mode 100644 problems/二叉树的理论基础.md
delete mode 100644 problems/背包理论基础01背包-一维DP.md
delete mode 100644 problems/背包理论基础01背包-二维DP.md
diff --git a/README.md b/README.md
index b4e82eb0..58f71049 100644
--- a/README.md
+++ b/README.md
@@ -12,10 +12,10 @@
-
+
-
+
# LeetCode 刷题攻略
@@ -302,7 +302,7 @@
动态规划专题已经开始啦,来不及解释了,小伙伴们上车别掉队!
-
+
1. [关于动态规划,你该了解这些!](./problems/动态规划理论基础.md)
2. [动态规划:斐波那契数](./problems/0509.斐波那契数.md)
3. [动态规划:爬楼梯](./problems/0070.爬楼梯.md)
diff --git a/problems/0018.四数之和.md b/problems/0018.四数之和.md
index a1591736..0caf12be 100644
--- a/problems/0018.四数之和.md
+++ b/problems/0018.四数之和.md
@@ -31,38 +31,39 @@ https://leetcode-cn.com/problems/4sum/
# 思路
-四数之和,和[三数之和](https://mp.weixin.qq.com/s/r5cgZFu0tv4grBAexdcd8A)是一个思路,都是使用双指针法, 基本解法就是在[三数之和](https://mp.weixin.qq.com/s/r5cgZFu0tv4grBAexdcd8A) 的基础上再套一层for循环。
+四数之和,和[15.三数之和](https://mp.weixin.qq.com/s/QfTNEByq1YlNSXRKEumwHg)是一个思路,都是使用双指针法, 基本解法就是在[15.三数之和](https://mp.weixin.qq.com/s/QfTNEByq1YlNSXRKEumwHg) 的基础上再套一层for循环。
但是有一些细节需要注意,例如: 不要判断`nums[k] > target` 就返回了,三数之和 可以通过 `nums[i] > 0` 就返回了,因为 0 已经是确定的数了,四数之和这道题目 target是任意值。(大家亲自写代码就能感受出来)
-[三数之和](https://mp.weixin.qq.com/s/r5cgZFu0tv4grBAexdcd8A)的双指针解法是一层for循环num[i]为确定值,然后循环内有left和right下表作为双指针,找到nums[i] + nums[left] + nums[right] == 0。
+[15.三数之和](https://mp.weixin.qq.com/s/QfTNEByq1YlNSXRKEumwHg)的双指针解法是一层for循环num[i]为确定值,然后循环内有left和right下表作为双指针,找到nums[i] + nums[left] + nums[right] == 0。
四数之和的双指针解法是两层for循环nums[k] + nums[i]为确定值,依然是循环内有left和right下表作为双指针,找出nums[k] + nums[i] + nums[left] + nums[right] == target的情况,三数之和的时间复杂度是O(n^2),四数之和的时间复杂度是O(n^3) 。
那么一样的道理,五数之和、六数之和等等都采用这种解法。
-对于[三数之和](https://mp.weixin.qq.com/s/r5cgZFu0tv4grBAexdcd8A)双指针法就是将原本暴力O(n^3)的解法,降为O(n^2)的解法,四数之和的双指针解法就是将原本暴力O(n^4)的解法,降为O(n^3)的解法。
+对于[15.三数之和](https://mp.weixin.qq.com/s/QfTNEByq1YlNSXRKEumwHg)双指针法就是将原本暴力O(n^3)的解法,降为O(n^2)的解法,四数之和的双指针解法就是将原本暴力O(n^4)的解法,降为O(n^3)的解法。
-之前我们讲过哈希表的经典题目:[四数相加II](https://mp.weixin.qq.com/s/Ue8pKKU5hw_m-jPgwlHcbA),相对于本题简单很多,因为本题是要求在一个集合中找出四个数相加等于target,同时四元组不能重复。
+之前我们讲过哈希表的经典题目:[454.四数相加II](https://mp.weixin.qq.com/s/12g_w6RzHuEpFts1pT6BWw),相对于本题简单很多,因为本题是要求在一个集合中找出四个数相加等于target,同时四元组不能重复。
-而[四数相加II](https://mp.weixin.qq.com/s/Ue8pKKU5hw_m-jPgwlHcbA)是四个独立的数组,只要找到A[i] + B[j] + C[k] + D[l] = 0就可以,不用考虑有重复的四个元素相加等于0的情况,所以相对于本题还是简单了不少!
+而[454.四数相加II](https://mp.weixin.qq.com/s/12g_w6RzHuEpFts1pT6BWw)是四个独立的数组,只要找到A[i] + B[j] + C[k] + D[l] = 0就可以,不用考虑有重复的四个元素相加等于0的情况,所以相对于本题还是简单了不少!
我们来回顾一下,几道题目使用了双指针法。
双指针法将时间复杂度O(n^2)的解法优化为 O(n)的解法。也就是降一个数量级,题目如下:
-* [0027.移除元素](https://mp.weixin.qq.com/s/wj0T-Xs88_FHJFwayElQlA)
-* [15.三数之和](https://mp.weixin.qq.com/s/r5cgZFu0tv4grBAexdcd8A)
+
+* [27.移除元素](https://mp.weixin.qq.com/s/RMkulE4NIb6XsSX83ra-Ww)
+* [15.三数之和](https://mp.weixin.qq.com/s/QfTNEByq1YlNSXRKEumwHg)
* [18.四数之和](https://mp.weixin.qq.com/s/nQrcco8AZJV1pAOVjeIU_g)
-双指针来记录前后指针实现链表反转:
-* [206.反转链表](https://mp.weixin.qq.com/s/pnvVP-0ZM7epB8y3w_Njwg)
+操作链表:
-使用双指针来确定有环:
+* [206.反转链表](https://mp.weixin.qq.com/s/ckEvIVGcNLfrz6OLOMoT0A)
+* [19.删除链表的倒数第N个节点](https://mp.weixin.qq.com/s/gxu65X1343xW_sBrkTz0Eg)
+* [面试题 02.07. 链表相交](https://mp.weixin.qq.com/s/BhfFfaGvt9Zs7UmH4YehZw)
+* [142题.环形链表II](https://mp.weixin.qq.com/s/gt_VH3hQTqNxyWcl1ECSbQ)
-* [142题.环形链表II](https://mp.weixin.qq.com/s/_QVP3IkRZWx9zIpQRgajzA)
-
-双指针法在数组和链表中还有很多应用,后面还会介绍到。
+双指针法在字符串题目中还有很多应用,后面还会介绍到。
C++代码
diff --git a/problems/0028.实现strStr.md b/problems/0028.实现strStr.md
index b8ebcaa1..6f257d5e 100644
--- a/problems/0028.实现strStr.md
+++ b/problems/0028.实现strStr.md
@@ -61,11 +61,6 @@ KMP的经典思想就是:**当出现字符串不匹配时,可以记录一部
读完本篇可以顺便,把leetcode上28.实现strStr()题目做了。
-如果文字实在看不下去,就看我在B站上的视频吧,如下:
-
-* [帮你把KMP算法学个通透!(理论篇)B站](https://www.bilibili.com/video/BV1PD4y1o7nd/)
-* [帮你把KMP算法学个通透!(求next数组代码篇)B站](https://www.bilibili.com/video/BV1M5411j7Xx/)
-
# 什么是KMP
diff --git a/problems/0454.四数相加II.md b/problems/0454.四数相加II.md
index 2c648f58..939ed20d 100644
--- a/problems/0454.四数相加II.md
+++ b/problems/0454.四数相加II.md
@@ -11,6 +11,7 @@
# 第454题.四数相加II
+
https://leetcode-cn.com/problems/4sum-ii/
给定四个包含整数的数组列表 A , B , C , D ,计算有多少个元组 (i, j, k, l) ,使得 A[i] + B[j] + C[k] + D[l] = 0。
@@ -24,10 +25,8 @@ A = [ 1, 2]
B = [-2,-1]
C = [-1, 2]
D = [ 0, 2]
-
输出:
2
-
**解释:**
两个元组如下:
1. (0, 0, 0, 1) -> A[0] + B[0] + C[0] + D[1] = 1 + (-2) + (-1) + 2 = 0
diff --git a/problems/二叉树的理论基础.md b/problems/二叉树的理论基础.md
deleted file mode 100644
index 4b52e511..00000000
--- a/problems/二叉树的理论基础.md
+++ /dev/null
@@ -1,202 +0,0 @@
-
-
-
-
-
-
-欢迎大家参与本项目,贡献其他语言版本的代码,拥抱开源,让更多学习算法的小伙伴们收益!
-
-## 二叉树理论基础
-
-我们要开启新的征程了,大家跟上!
-
-说道二叉树,大家对于二叉树其实都很熟悉了,本文呢我也不想教科书式的把二叉树的基础内容再啰嗦一遍,所以一下我讲的都是一些比较重点的内容。
-
-相信只要耐心看完,都会有所收获。
-
-## 二叉树的种类
-
-在我们解题过程中二叉树有两种主要的形式:满二叉树和完全二叉树。
-
-### 满二叉树
-
-满二叉树:如果一棵二叉树只有度为0的结点和度为2的结点,并且度为0的结点在同一层上,则这棵二叉树为满二叉树。
-
-如图所示:
-
-
-
-这棵二叉树为满二叉树,也可以说深度为 k,有 $(2^k)-1$ 个节点的二叉树。
-
-
-### 完全二叉树
-
-什么是完全二叉树?
-
-完全二叉树的定义如下:在完全二叉树中,除了最底层节点可能没填满外,其余每层节点数都达到最大值,并且最下面一层的节点都集中在该层最左边的若干位置。若最底层为第 h 层,则该层包含 1 ~ $2^{(h-1)}$ 个节点。
-
-**大家要自己看完全二叉树的定义,很多同学对完全二叉树其实不是真正的懂了。**
-
-我来举一个典型的例子如题:
-
-
-
-相信不少同学最后一个二叉树是不是完全二叉树都中招了。
-
-**之前我们刚刚讲过优先级队列其实是一个堆,堆就是一棵完全二叉树,同时保证父子节点的顺序关系。**
-
-### 二叉搜索树
-
-前面介绍的书,都没有数值的,而二叉搜索树是有数值的了,**二叉搜索树是一个有序树**。
-
-
-* 若它的左子树不空,则左子树上所有结点的值均小于它的根结点的值;
-* 若它的右子树不空,则右子树上所有结点的值均大于它的根结点的值;
-* 它的左、右子树也分别为二叉排序树
-
-下面这两棵树都是搜索树
-
-
-
-### 平衡二叉搜索树
-
-平衡二叉搜索树:又被称为AVL(Adelson-Velsky and Landis)树,且具有以下性质:它是一棵空树或它的左右两个子树的高度差的绝对值不超过1,并且左右两个子树都是一棵平衡二叉树。
-
-如图:
-
-
-
-最后一棵 不是平衡二叉树,因为它的左右两个子树的高度差的绝对值超过了1。
-
-**C++中map、set、multimap,multiset的底层实现都是平衡二叉搜索树**,所以map、set的增删操作时间时间复杂度是logn,注意我这里没有说unordered_map、unordered_set,unordered_map、unordered_map底层实现是哈希表。
-
-**所以大家使用自己熟悉的编程语言写算法,一定要知道常用的容器底层都是如何实现的,最基本的就是map、set等等,否则自己写的代码,自己对其性能分析都分析不清楚!**
-
-
-## 二叉树的存储方式
-
-**二叉树可以链式存储,也可以顺序存储。**
-
-那么链式存储方式就用指针, 顺序存储的方式就是用数组。
-
-顾名思义就是顺序存储的元素在内存是连续分布的,而链式存储则是通过指针把分布在散落在各个地址的节点串联一起。
-
-链式存储如图:
-
-
-
-链式存储是大家很熟悉的一种方式,那么我们来看看如何顺序存储呢?
-
-其实就是用数组来存储二叉树,顺序存储的方式如图:
-
-
-
-用数组来存储二叉树如何遍历的呢?
-
-**如果父节点的数组下标是 i,那么它的左孩子就是 i * 2 + 1,右孩子就是 i * 2 + 2。**
-
-但是用链式表示的二叉树,更有利于我们理解,所以一般我们都是用链式存储二叉树。
-
-**所以大家要了解,用数组依然可以表示二叉树。**
-
-## 二叉树的遍历方式
-
-关于二叉树的遍历方式,要知道二叉树遍历的基本方式都有哪些。
-
-一些同学用做了很多二叉树的题目了,可能知道前序、中序、后序遍历,可能知道层序遍历,但是却没有框架。
-
-我这里把二叉树的几种遍历方式列出来,大家就可以一一串起来了。
-
-二叉树主要有两种遍历方式:
-1. 深度优先遍历:先往深走,遇到叶子节点再往回走。
-2. 广度优先遍历:一层一层的去遍历。
-
-**这两种遍历是图论中最基本的两种遍历方式**,后面在介绍图论的时候 还会介绍到。
-
-那么从深度优先遍历和广度优先遍历进一步拓展,才有如下遍历方式:
-
-* 深度优先遍历
- * 前序遍历(递归法,迭代法)
- * 中序遍历(递归法,迭代法)
- * 后序遍历(递归法,迭代法)
-* 广度优先遍历
- * 层次遍历(迭代法)
-
-
-在深度优先遍历中:有三个顺序,前序、中序、后序遍历, 有同学总分不清这三个顺序,经常搞混,我这里教大家一个技巧。
-
-**这里前、中、后,其实指的就是中间节点的遍历顺序**,只要大家记住 前序、中序、后序指的就是中间节点的位置就可以了。
-
-看如下中间节点的顺序,就可以发现,中间节点的顺序就是所谓的遍历方式
-
-* 前序遍历:中左右
-* 中序遍历:左中右
-* 后序遍历:左右中
-
-大家可以对着如下图,看看自己理解的前后中序有没有问题。
-
-
-
-最后再说一说二叉树中深度优先和广度优先遍历实现方式,我们做二叉树相关题目,经常会使用递归的方式来实现深度优先遍历,也就是实现前序、中序、后序遍历,使用递归是比较方便的。
-
-**之前我们讲栈与队列的时候,就说过栈其实就是递归的一种是实现结构**,也就说前序、中序、后序遍历的逻辑其实都是可以借助栈使用非递归的方式来实现的。
-
-而广度优先遍历的实现一般使用队列来实现,这也是队列先进先出的特点所决定的,因为需要先进先出的结构,才能一层一层的来遍历二叉树。
-
-**这里其实我们又了解了栈与队列的一个应用场景了。**
-
-具体的实现我们后面都会讲的,这里大家先要清楚这些理论基础。
-
-## 二叉树的定义
-
-刚刚我们说过了二叉树有两种存储方式顺序存储,和链式存储,顺序存储就是用数组来存,这个定义没啥可说的,我们来看看链式存储的二叉树节点的定义方式。
-
-
-C++代码如下:
-
-```
-struct TreeNode {
- int val;
- TreeNode *left;
- TreeNode *right;
- TreeNode(int x) : val(x), left(NULL), right(NULL) {}
-};
-```
-
-大家会发现二叉树的定义 和链表是差不多的,相对于链表 ,二叉树的节点里多了一个指针, 有两个指针,指向左右孩子.
-
-这里要提醒大家要注意二叉树节点定义的书写方式。
-
-**在现场面试的时候 面试官可能要求手写代码,所以数据结构的定义以及简单逻辑的代码一定要锻炼白纸写出来。**
-
-因为我们在刷leetcode的时候,节点的定义默认都定义好了,真到面试的时候,需要自己写节点定义的时候,有时候会一脸懵逼!
-
-## 总结
-
-二叉树是一种基础数据结构,在算法面试中都是常客,也是众多数据结构的基石。
-
-本篇我们介绍了二叉树的种类、存储方式、遍历方式以及定义,比较全面的介绍了二叉树各个方面的重点,帮助大家扫一遍基础。
-
-**说道二叉树,就不得不说递归,很多同学对递归都是又熟悉又陌生,递归的代码一般很简短,但每次都是一看就会,一写就废。**
-
-
-
-## 其他语言版本
-
-
-Java:
-
-
-Python:
-
-
-Go:
-
-
-
-
------------------------
-* 作者微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
-* B站视频:[代码随想录](https://space.bilibili.com/525438321)
-* 知识星球:[代码随想录](https://mp.weixin.qq.com/s/QVF6upVMSbgvZy8lHZS3CQ)
-
diff --git a/problems/背包理论基础01背包-一维DP.md b/problems/背包理论基础01背包-一维DP.md
deleted file mode 100644
index 47f489ea..00000000
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+++ /dev/null
@@ -1,231 +0,0 @@
-
-
-
-
-
-
-欢迎大家参与本项目,贡献其他语言版本的代码,拥抱开源,让更多学习算法的小伙伴们收益!
-
-# 动态规划:关于01背包问题,你该了解这些!(滚动数组)
-
-昨天[动态规划:关于01背包问题,你该了解这些!](https://mp.weixin.qq.com/s/FwIiPPmR18_AJO5eiidT6w)中是用二维dp数组来讲解01背包。
-
-今天我们就来说一说滚动数组,其实在前面的题目中我们已经用到过滚动数组了,就是把二维dp降为一维dp,一些录友当时还表示比较困惑。
-
-那么我们通过01背包,来彻底讲一讲滚动数组!
-
-接下来还是用如下这个例子来进行讲解
-
-背包最大重量为4。
-
-物品为:
-
-| | 重量 | 价值 |
-| --- | --- | --- |
-| 物品0 | 1 | 15 |
-| 物品1 | 3 | 20 |
-| 物品2 | 4 | 30 |
-
-问背包能背的物品最大价值是多少?
-
-## 一维dp数组(滚动数组)
-
-对于背包问题其实状态都是可以压缩的。
-
-在使用二维数组的时候,递推公式:dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - weight[i]] + value[i]);
-
-**其实可以发现如果把dp[i - 1]那一层拷贝到dp[i]上,表达式完全可以是:dp[i][j] = max(dp[i][j], dp[i][j - weight[i]] + value[i]);**
-
-**于其把dp[i - 1]这一层拷贝到dp[i]上,不如只用一个一维数组了**,只用dp[j](一维数组,也可以理解是一个滚动数组)。
-
-这就是滚动数组的由来,需要满足的条件是上一层可以重复利用,直接拷贝到当前层。
-
-读到这里估计大家都忘了 dp[i][j]里的i和j表达的是什么了,i是物品,j是背包容量。
-
-**dp[i][j] 表示从下标为[0-i]的物品里任意取,放进容量为j的背包,价值总和最大是多少**。
-
-一定要时刻记住这里i和j的含义,要不然很容易看懵了。
-
-动规五部曲分析如下:
-
-1. 确定dp数组的定义
-
-在一维dp数组中,dp[j]表示:容量为j的背包,所背的物品价值可以最大为dp[j]。
-
-2. 一维dp数组的递推公式
-
-dp[j]为 容量为j的背包所背的最大价值,那么如何推导dp[j]呢?
-
-dp[j]可以通过dp[j - weight[j]]推导出来,dp[j - weight[i]]表示容量为j - weight[i]的背包所背的最大价值。
-
-dp[j - weight[i]] + value[i] 表示 容量为 j - 物品i重量 的背包 加上 物品i的价值。(也就是容量为j的背包,放入物品i了之后的价值即:dp[j])
-
-此时dp[j]有两个选择,一个是取自己dp[j],一个是取dp[j - weight[i]] + value[i],指定是取最大的,毕竟是求最大价值,
-
-所以递归公式为:
-
-```
-dp[j] = max(dp[j], dp[j - weight[i]] + value[i]);
-```
-
-可以看出相对于二维dp数组的写法,就是把dp[i][j]中i的维度去掉了。
-
-3. 一维dp数组如何初始化
-
-**关于初始化,一定要和dp数组的定义吻合,否则到递推公式的时候就会越来越乱**。
-
-dp[j]表示:容量为j的背包,所背的物品价值可以最大为dp[j],那么dp[0]就应该是0,因为背包容量为0所背的物品的最大价值就是0。
-
-那么dp数组除了下标0的位置,初始为0,其他下标应该初始化多少呢?
-
-看一下递归公式:dp[j] = max(dp[j], dp[j - weight[i]] + value[i]);
-
-dp数组在推导的时候一定是取价值最大的数,如果题目给的价值都是正整数那么非0下标都初始化为0就可以了,如果题目给的价值有负数,那么非0下标就要初始化为负无穷。
-
-**这样才能让dp数组在递归公式的过程中取的最大的价值,而不是被初始值覆盖了**。
-
-那么我假设物品价值都是大于0的,所以dp数组初始化的时候,都初始为0就可以了。
-
-4. 一维dp数组遍历顺序
-
-代码如下:
-
-```
-for(int i = 0; i < weight.size(); i++) { // 遍历物品
- for(int j = bagWeight; j >= weight[i]; j--) { // 遍历背包容量
- dp[j] = max(dp[j], dp[j - weight[i]] + value[i]);
-
- }
-}
-```
-
-**这里大家发现和二维dp的写法中,遍历背包的顺序是不一样的!**
-
-二维dp遍历的时候,背包容量是从小到大,而一维dp遍历的时候,背包是从大到小。
-
-为什么呢?
-
-**倒叙遍历是为了保证物品i只被放入一次!**,在[动态规划:关于01背包问题,你该了解这些!](https://mp.weixin.qq.com/s/FwIiPPmR18_AJO5eiidT6w)中讲解二维dp数组初始化dp[0][j]时候已经讲解到过一次。
-
-举一个例子:物品0的重量weight[0] = 1,价值value[0] = 15
-
-如果正序遍历
-
-dp[1] = dp[1 - weight[0]] + value[0] = 15
-
-dp[2] = dp[2 - weight[0]] + value[0] = 30
-
-此时dp[2]就已经是30了,意味着物品0,被放入了两次,所以不能正序遍历。
-
-为什么倒叙遍历,就可以保证物品只放入一次呢?
-
-倒叙就是先算dp[2]
-
-dp[2] = dp[2 - weight[0]] + value[0] = 15 (dp数组已经都初始化为0)
-
-dp[1] = dp[1 - weight[0]] + value[0] = 15
-
-所以从后往前循环,每次取得状态不会和之前取得状态重合,这样每种物品就只取一次了。
-
-**那么问题又来了,为什么二维dp数组历的时候不用倒叙呢?**
-
-因为对于二维dp,dp[i][j]都是通过上一层即dp[i - 1][j]计算而来,本层的dp[i][j]并不会被覆盖!
-
-(如何这里读不懂,大家就要动手试一试了,空想还是不靠谱的,实践出真知!)
-
-**再来看看两个嵌套for循环的顺序,代码中是先遍历物品嵌套遍历背包容量,那可不可以先遍历背包容量嵌套遍历物品呢?**
-
-不可以!
-
-因为一维dp的写法,背包容量一定是要倒序遍历(原因上面已经讲了),如果遍历背包容量放在上一层,那么每个dp[j]就只会放入一个物品,即:背包里只放入了一个物品。
-
-(这里如果读不懂,就在回想一下dp[j]的定义,或者就把两个for循环顺序颠倒一下试试!)
-
-**所以一维dp数组的背包在遍历顺序上和二维其实是有很大差异的!**,这一点大家一定要注意。
-
-5. 举例推导dp数组
-
-一维dp,分别用物品0,物品1,物品2 来遍历背包,最终得到结果如下:
-
-
-
-
-
-## 一维dp01背包完整C++测试代码
-
-```
-void test_1_wei_bag_problem() {
- vector weight = {1, 3, 4};
- vector value = {15, 20, 30};
- int bagWeight = 4;
-
- // 初始化
- vector dp(bagWeight + 1, 0);
- for(int i = 0; i < weight.size(); i++) { // 遍历物品
- for(int j = bagWeight; j >= weight[i]; j--) { // 遍历背包容量
- dp[j] = max(dp[j], dp[j - weight[i]] + value[i]);
- }
- }
- cout << dp[bagWeight] << endl;
-}
-
-int main() {
- test_1_wei_bag_problem();
-}
-
-```
-
-可以看出,一维dp 的01背包,要比二维简洁的多! 初始化 和 遍历顺序相对简单了。
-
-**所以我倾向于使用一维dp数组的写法,比较直观简洁,而且空间复杂度还降了一个数量级!**
-
-**在后面背包问题的讲解中,我都直接使用一维dp数组来进行推导**。
-
-## 总结
-
-以上的讲解可以开发一道面试题目(毕竟力扣上没原题)。
-
-就是本文中的题目,要求先实现一个纯二维的01背包,如果写出来了,然后再问为什么两个for循环的嵌套顺序这么写?反过来写行不行?再讲一讲初始化的逻辑。
-
-然后要求实现一个一维数组的01背包,最后再问,一维数组的01背包,两个for循环的顺序反过来写行不行?为什么?
-
-注意以上问题都是在候选人把代码写出来的情况下才问的。
-
-就是纯01背包的题目,都不用考01背包应用类的题目就可以看出候选人对算法的理解程度了。
-
-**相信大家读完这篇文章,应该对以上问题都有了答案!**
-
-此时01背包理论基础就讲完了,我用了两篇文章把01背包的dp数组定义、递推公式、初始化、遍历顺序从二维数组到一维数组统统深度剖析了一遍,没有放过任何难点。
-
-大家可以发现其实信息量还是挺大的。
-
-如果把[动态规划:关于01背包问题,你该了解这些!](https://mp.weixin.qq.com/s/FwIiPPmR18_AJO5eiidT6w)和本篇的内容都理解了,后面我们在做01背包的题目,就会发现非常简单了。
-
-不用再凭感觉或者记忆去写背包,而是有自己的思考,了解其本质,代码的方方面面都在自己的掌控之中。
-
-即使代码没有通过,也会有自己的逻辑去debug,这样就思维清晰了。
-
-接下来就要开始用这两天的理论基础去做力扣上的背包面试题目了,录友们握紧扶手,我们要上高速啦!
-
-就酱,学算法,认准「代码随想录」,值得你推荐给身边每一位朋友同学们。
-
-
-## 其他语言版本
-
-
-Java:
-
-
-Python:
-
-
-Go:
-
-
-
-
------------------------
-* 作者微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
-* B站视频:[代码随想录](https://space.bilibili.com/525438321)
-* 知识星球:[代码随想录](https://mp.weixin.qq.com/s/QVF6upVMSbgvZy8lHZS3CQ)
-
diff --git a/problems/背包理论基础01背包-二维DP.md b/problems/背包理论基础01背包-二维DP.md
deleted file mode 100644
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--- a/problems/背包理论基础01背包-二维DP.md
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@@ -1,318 +0,0 @@
-
-
-
-
-
-
-欢迎大家参与本项目,贡献其他语言版本的代码,拥抱开源,让更多学习算法的小伙伴们收益!
-
-# 背包问题理论基础
-
-
-这周我们正式开始讲解背包问题!
-
-背包问题的经典资料当然是:背包九讲。在公众号「代码随想录」后台回复:背包九讲,就可以获得背包九讲的PDF。
-
-但说实话,背包九讲对于小白来说确实不太友好,看起来还是有点费劲的,而且都是伪代码理解起来也吃力。
-
-对于面试的话,其实掌握01背包,和完全背包,就够用了,最多可以再来一个多重背包。
-
-如果这几种背包,分不清,我这里画了一个图,如下:
-
-
-
-至于背包九讲其其他背包,面试几乎不会问,都是竞赛级别的了,leetcode上连多重背包的题目都没有,所以题库也告诉我们,01背包和完全背包就够用了。
-
-而完全背包又是也是01背包稍作变化而来,即:完全背包的物品数量是无限的。
-
-**所以背包问题的理论基础重中之重是01背包,一定要理解透!**
-
-leetcode上没有纯01背包的问题,都是01背包应用方面的题目,也就是需要转化为01背包问题。
-
-**所以我先通过纯01背包问题,把01背包原理讲清楚,后续再讲解leetcode题目的时候,重点就是讲解如何转化为01背包问题了**。
-
-之前可能有些录友已经可以熟练写出背包了,但只要把这个文章仔细看完,相信你会意外收获!
-
-## 01 背包
-
-有N件物品和一个最多能被重量为W 的背包。第i件物品的重量是weight[i],得到的价值是value[i] 。**每件物品只能用一次**,求解将哪些物品装入背包里物品价值总和最大。
-
-
-
-这是标准的背包问题,以至于很多同学看了这个自然就会想到背包,甚至都不知道暴力的解法应该怎么解了。
-
-这样其实是没有从底向上去思考,而是习惯性想到了背包,那么暴力的解法应该是怎么样的呢?
-
-每一件物品其实只有两个状态,取或者不取,所以可以使用回溯法搜索出所有的情况,那么时间复杂度就是O(2^n),这里的n表示物品数量。
-
-**所以暴力的解法是指数级别的时间复杂度。进而才需要动态规划的解法来进行优化!**
-
-在下面的讲解中,我举一个例子:
-
-背包最大重量为4。
-
-物品为:
-
-| | 重量 | 价值 |
-| --- | --- | --- |
-| 物品0 | 1 | 15 |
-| 物品1 | 3 | 20 |
-| 物品2 | 4 | 30 |
-
-问背包能背的物品最大价值是多少?
-
-以下讲解和图示中出现的数字都是以这个例子为例。
-
-## 二维dp数组01背包
-
-依然动规五部曲分析一波。
-
-1. 确定dp数组以及下标的含义
-
-对于背包问题,有一种写法, 是使用二维数组,即**dp[i][j] 表示从下标为[0-i]的物品里任意取,放进容量为j的背包,价值总和最大是多少**。
-
-只看这个二维数组的定义,大家一定会有点懵,看下面这个图:
-
-
-
-**要时刻记着这个dp数组的含义,下面的一些步骤都围绕这dp数组的含义进行的**,如果哪里看懵了,就来回顾一下i代表什么,j又代表什么。
-
-2. 确定递推公式
-
-再回顾一下dp[i][j]的含义:从下标为[0-i]的物品里任意取,放进容量为j的背包,价值总和最大是多少。
-
-那么可以有两个方向推出来dp[i][j],
-
-* 由dp[i - 1][j]推出,即背包容量为j,里面不放物品i的最大价值,此时dp[i][j]就是dp[i - 1][j]
-* 由dp[i - 1][j - weight[i]]推出,dp[i - 1][j - weight[i]] 为背包容量为j - weight[i]的时候不放物品i的最大价值,那么dp[i - 1][j - weight[i]] + value[i] (物品i的价值),就是背包放物品i得到的最大价值
-
-所以递归公式: dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - weight[i]] + value[i]);
-
-3. dp数组如何初始化
-
-**关于初始化,一定要和dp数组的定义吻合,否则到递推公式的时候就会越来越乱**。
-
-首先从dp[i][j]的定义触发,如果背包容量j为0的话,即dp[i][0],无论是选取哪些物品,背包价值总和一定为0。如图:
-
-
-
-在看其他情况。
-
-状态转移方程 dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - weight[i]] + value[i]); 可以看出i 是由 i-1 推导出来,那么i为0的时候就一定要初始化。
-
-dp[0][j],即:i为0,存放编号0的物品的时候,各个容量的背包所能存放的最大价值。
-
-代码如下:
-
-```
-// 倒叙遍历
-for (int j = bagWeight; j >= weight[0]; j--) {
- dp[0][j] = dp[0][j - weight[0]] + value[0]; // 初始化i为0时候的情况
-}
-```
-
-**大家应该发现,这个初始化为什么是倒叙的遍历的?正序遍历就不行么?**
-
-正序遍历还真就不行,dp[0][j]表示容量为j的背包存放物品0时候的最大价值,物品0的价值就是15,因为题目中说了**每个物品只有一个!**所以dp[0][j]如果不是初始值的话,就应该都是物品0的价值,也就是15。
-
-但如果一旦正序遍历了,那么物品0就会被重复加入多次! 例如代码如下:
-```
-// 正序遍历
-for (int j = weight[0]; j <= bagWeight; j++) {
- dp[0][j] = dp[0][j - weight[0]] + value[0];
-}
-```
-
-例如dp[0][1] 是15,到了dp[0][2] = dp[0][2 - 1] + 15; 也就是dp[0][2] = 30 了,那么就是物品0被重复放入了。
-
-**所以一定要倒叙遍历,保证物品0只被放入一次!这一点对01背包很重要,后面在讲解滚动数组的时候,还会用到倒叙遍历来保证物品使用一次!**
-
-
-此时dp数组初始化情况如图所示:
-
-
-
-dp[0][j] 和 dp[i][0] 都已经初始化了,那么其他下标应该初始化多少呢?
-
-
-dp[i][j]在推导的时候一定是取价值最大的数,如果题目给的价值都是正整数那么非0下标都初始化为0就可以了,因为0就是最小的了,不会影响取最大价值的结果。
-
-如果题目给的价值有负数,那么非0下标就要初始化为负无穷了。例如:一个物品的价值是-2,但对应的位置依然初始化为0,那么取最大值的时候,就会取0而不是-2了,所以要初始化为负无穷。
-
-**这样才能让dp数组在递归公式的过程中取最大的价值,而不是被初始值覆盖了**。
-
-最后初始化代码如下:
-
-```
-// 初始化 dp
-vector> dp(weight.size() + 1, vector(bagWeight + 1, 0));
-for (int j = bagWeight; j >= weight[0]; j--) {
- dp[0][j] = dp[0][j - weight[0]] + value[0];
-}
-```
-
-**费了这么大的功夫,才把如何初始化讲清楚,相信不少同学平时初始化dp数组是凭感觉来的,但有时候感觉是不靠谱的**。
-
-4. 确定遍历顺序
-
-
-在如下图中,可以看出,有两个遍历的维度:物品与背包重量
-
-
-
-那么问题来了,**先遍历 物品还是先遍历背包重量呢?**
-
-**其实都可以!! 但是先遍历物品更好理解**。
-
-那么我先给出先遍历物品,然后遍历背包重量的代码。
-
-```
-// weight数组的大小 就是物品个数
-for(int i = 1; i < weight.size(); i++) { // 遍历物品
- for(int j = 0; j <= bagWeight; j++) { // 遍历背包容量
- if (j < weight[i]) dp[i][j] = dp[i - 1][j]; // 这个是为了展现dp数组里元素的变化
- else dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - weight[i]] + value[i]);
-
- }
-}
-```
-
-**先遍历背包,再遍历物品,也是可以的!(注意我这里使用的二维dp数组)**
-
-例如这样:
-
-```
-// weight数组的大小 就是物品个数
-for(int j = 0; j <= bagWeight; j++) { // 遍历背包容量
- for(int i = 1; i < weight.size(); i++) { // 遍历物品
- if (j < weight[i]) dp[i][j] = dp[i - 1][j];
- else dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - weight[i]] + value[i]);
- }
-}
-```
-
-为什么也是可以的呢?
-
-**要理解递归的本质和递推的方向**。
-
-dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - weight[i]] + value[i]); 递归公式中可以看出dp[i][j]是靠dp[i-1][j]和dp[i - 1][j - weight[i]]推导出来的。
-
-dp[i-1][j]和dp[i - 1][j - weight[i]] 都在dp[i][j]的左上角方向(包括正左和正上两个方向),那么先遍历物品,再遍历背包的过程如图所示:
-
-
-
-再来看看先遍历背包,再遍历物品呢,如图:
-
-
-
-**大家可以看出,虽然两个for循环遍历的次序不同,但是dp[i][j]所需要的数据就是左上角,根本不影响dp[i][j]公式的推导!**
-
-但先遍历物品再遍历背包这个顺序更好理解。
-
-**其实背包问题里,两个for循环的先后循序是非常有讲究的,理解遍历顺序其实比理解推导公式难多了**。
-
-5. 举例推导dp数组
-
-来看一下对应的dp数组的数值,如图:
-
-
-
-最终结果就是dp[2][4]。
-
-建议大家此时自己在纸上推导一遍,看看dp数组里每一个数值是不是这样的。
-
-**做动态规划的题目,最好的过程就是自己在纸上举一个例子把对应的dp数组的数值推导一下,然后在动手写代码!**
-
-很多同学做dp题目,遇到各种问题,然后凭感觉东改改西改改,怎么改都不对,或者稀里糊涂就改过了。
-
-主要就是自己没有动手推导一下dp数组的演变过程,如果推导明白了,代码写出来就算有问题,只要把dp数组打印出来,对比一下和自己推导的有什么差异,很快就可以发现问题了。
-
-
-## 完整C++测试代码
-
-```C++
-void test_2_wei_bag_problem1() {
- vector weight = {1, 3, 4};
- vector value = {15, 20, 30};
- int bagWeight = 4;
-
- // 二维数组
- vector> dp(weight.size() + 1, vector(bagWeight + 1, 0));
-
- // 初始化
- for (int j = bagWeight; j >= weight[0]; j--) {
- dp[0][j] = dp[0][j - weight[0]] + value[0];
- }
-
- // weight数组的大小 就是物品个数
- for(int i = 1; i < weight.size(); i++) { // 遍历物品
- for(int j = 0; j <= bagWeight; j++) { // 遍历背包容量
- if (j < weight[i]) dp[i][j] = dp[i - 1][j];
- else dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - weight[i]] + value[i]);
-
- }
- }
-
- cout << dp[weight.size() - 1][bagWeight] << endl;
-}
-
-int main() {
- test_2_wei_bag_problem1();
-}
-
-```
-
-
-以上遍历的过程也可以这么写:
-
-```
-// 遍历过程
-for(int i = 1; i < weight.size(); i++) { // 遍历物品
- for(int j = 0; j <= bagWeight; j++) { // 遍历背包容量
- if (j - weight[i] >= 0) {
- dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - weight[i]] + value[i]);
- }
- }
-}
-```
-
-这么写打印出来的dp数据这就是这样:
-
-
-
-空出来的0其实是用不上的,版本一 能把完整的dp数组打印出来,出来我用版本一来讲解。
-
-
-## 总结
-
-讲了这么多才刚刚把二维dp的01背包讲完,**这里大家其实可以发现最简单的是推导公式了,推导公式估计看一遍就记下来了,但难就难在如何初始化和遍历顺序上**。
-
-可能有的同学并没有注意到初始化 和 遍历顺序的重要性,我们后面做力扣上背包面试题目的时候,大家就会感受出来了。
-
-下一篇 还是理论基础,我们再来讲一维dp数组实现的01背包(滚动数组),分析一下和二维有什么区别,在初始化和遍历顺序上又有什么差异,敬请期待!
-
-就酱,学算法,认准「代码随想录」,值得推荐给身边的朋友同学们,关注后都会发现相见恨晚!
-
-
-
-
-
-## 其他语言版本
-
-
-Java:
-
-
-Python:
-
-
-Go:
-
-
-
-
------------------------
-* 作者微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
-* B站视频:[代码随想录](https://space.bilibili.com/525438321)
-* 知识星球:[代码随想录](https://mp.weixin.qq.com/s/QVF6upVMSbgvZy8lHZS3CQ)
-
From a00f188ddb6e60fd5a7da0042fe20645da746228 Mon Sep 17 00:00:00 2001
From: youngyangyang04 <826123027@qq.com>
Date: Fri, 4 Jun 2021 09:33:18 +0800
Subject: [PATCH 75/95] =?UTF-8?q?=E6=9B=B4=E6=96=B0=E5=A4=B4=E9=83=A8?=
=?UTF-8?q?=E5=BA=95=E9=83=A8=E4=BF=A1=E6=81=AF?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
...试流程和注意事项都在这里了.md | 22 ++++++++++---------
...时了,此时的n究竟是多大?.md | 22 ++++++++++---------
.../前序/上海互联网公司总结.md | 20 +++++++++--------
...代码模式,什么又是ACM模式?.md | 19 ++++++++--------
problems/前序/代码风格.md | 21 +++++++++---------
...杂度,你不知道的都在这里!.md | 19 ++++++++--------
...间复杂度,可能有几个疑问?.md | 20 +++++++++--------
...了解自己代码的内存消耗么?.md | 22 ++++++++++---------
...上的代码想在本地编译运行?.md | 18 ++++++++-------
.../前序/北京互联网公司总结.md | 22 ++++++++++---------
.../前序/广州互联网公司总结.md | 20 +++++++++--------
.../前序/成都互联网公司总结.md | 22 ++++++++++---------
.../前序/杭州互联网公司总结.md | 22 ++++++++++---------
.../前序/深圳互联网公司总结.md | 20 +++++++++--------
problems/前序/程序员写文档工具.md | 22 ++++++++++---------
problems/前序/程序员简历.md | 17 +++++++-------
...算法的时间与空间复杂度分析.md | 20 +++++++++--------
...一讲递归算法的时间复杂度!.md | 20 +++++++++--------
18 files changed, 200 insertions(+), 168 deletions(-)
diff --git a/problems/前序/BAT级别技术面试流程和注意事项都在这里了.md b/problems/前序/BAT级别技术面试流程和注意事项都在这里了.md
index f8598d3f..fbcdf970 100644
--- a/problems/前序/BAT级别技术面试流程和注意事项都在这里了.md
+++ b/problems/前序/BAT级别技术面试流程和注意事项都在这里了.md
@@ -1,11 +1,12 @@
-
-
-
+
-
+
+欢迎大家参与本项目,贡献其他语言版本的代码,拥抱开源,让更多学习算法的小伙伴们收益!
+
+
大型互联网企业一般通过几轮技术面试来考察大家的各项能力,一般流程如下:
@@ -212,11 +213,12 @@ leetcode是专门针对算法练习的题库,leetcode现在也推出了中文
大家加油!
-------------------------
-* 微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
-* B站:[代码随想录](https://space.bilibili.com/525438321)
+
+
+
+-----------------------
+* 作者微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
+* B站视频:[代码随想录](https://space.bilibili.com/525438321)
* 知识星球:[代码随想录](https://mp.weixin.qq.com/s/QVF6upVMSbgvZy8lHZS3CQ)
-
-
-
+
diff --git a/problems/前序/On的算法居然超时了,此时的n究竟是多大?.md b/problems/前序/On的算法居然超时了,此时的n究竟是多大?.md
index f640abad..f6218777 100644
--- a/problems/前序/On的算法居然超时了,此时的n究竟是多大?.md
+++ b/problems/前序/On的算法居然超时了,此时的n究竟是多大?.md
@@ -1,11 +1,12 @@
-
-
-
+
-
+
+欢迎大家参与本项目,贡献其他语言版本的代码,拥抱开源,让更多学习算法的小伙伴们收益!
+
+
一些同学可能对计算机运行的速度还没有概念,就是感觉计算机运行速度应该会很快,那么在leetcode上做算法题目的时候为什么会超时呢?
@@ -219,11 +220,12 @@ int main() {
就酱,如果感觉「代码随想录」很干货,就帮忙宣传一波吧,很多录友发现这里之后都感觉相见恨晚!
-------------------------
-* 微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
-* B站:[代码随想录](https://space.bilibili.com/525438321)
+
+
+
+-----------------------
+* 作者微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
+* B站视频:[代码随想录](https://space.bilibili.com/525438321)
* 知识星球:[代码随想录](https://mp.weixin.qq.com/s/QVF6upVMSbgvZy8lHZS3CQ)
-
-
-
+
diff --git a/problems/前序/上海互联网公司总结.md b/problems/前序/上海互联网公司总结.md
index 40478cb6..386a0a93 100644
--- a/problems/前序/上海互联网公司总结.md
+++ b/problems/前序/上海互联网公司总结.md
@@ -1,10 +1,11 @@
-
-
+
-
+
+欢迎大家参与本项目,贡献其他语言版本的代码,拥抱开源,让更多学习算法的小伙伴们收益!
+
# 上海互联网公司总结
@@ -124,11 +125,12 @@
相对于北京和上海,深圳互联网公司断层很明显,腾讯一家独大,二线三线垂直行业的公司很少,所以说深圳腾讯的员工流动性相对是较低的,因为基本没得选。
-------------------------
-* 微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
-* B站:[代码随想录](https://space.bilibili.com/525438321)
+
+
+
+-----------------------
+* 作者微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
+* B站视频:[代码随想录](https://space.bilibili.com/525438321)
* 知识星球:[代码随想录](https://mp.weixin.qq.com/s/QVF6upVMSbgvZy8lHZS3CQ)
-
-
-
+
diff --git a/problems/前序/什么是核心代码模式,什么又是ACM模式?.md b/problems/前序/什么是核心代码模式,什么又是ACM模式?.md
index 70651817..5a19a72a 100644
--- a/problems/前序/什么是核心代码模式,什么又是ACM模式?.md
+++ b/problems/前序/什么是核心代码模式,什么又是ACM模式?.md
@@ -1,12 +1,12 @@
-
-
+
-
+
+欢迎大家参与本项目,贡献其他语言版本的代码,拥抱开源,让更多学习算法的小伙伴们收益!
+
---------------------------
现在很多企业都在牛客上进行面试,**很多录友和我反馈说搞不懂牛客上输入代码的ACM模式**。
什么是ACM输入模式呢? 就是自己构造输入数据格式,把要需要处理的容器填充好,OJ不会给你任何代码,包括include哪些函数都要自己写,最后也要自己控制返回数据的格式。
@@ -115,10 +115,11 @@ int main() {
如果大家有精力的话,也可以去POJ上去刷刷题,POJ是ACM选手首选OJ,输入模式也是ACM模式。
-------------------------
-* 微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
-* B站:[代码随想录](https://space.bilibili.com/525438321)
+
+
+-----------------------
+* 作者微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
+* B站视频:[代码随想录](https://space.bilibili.com/525438321)
* 知识星球:[代码随想录](https://mp.weixin.qq.com/s/QVF6upVMSbgvZy8lHZS3CQ)
-
-
+
diff --git a/problems/前序/代码风格.md b/problems/前序/代码风格.md
index e4378b4e..a9eabd0c 100644
--- a/problems/前序/代码风格.md
+++ b/problems/前序/代码风格.md
@@ -1,13 +1,13 @@
-
-
-
+
-
+
+欢迎大家参与本项目,贡献其他语言版本的代码,拥抱开源,让更多学习算法的小伙伴们收益!
+
+
---------------------------
# 看了这么多代码,谈一谈代码风格!
@@ -142,10 +142,11 @@ Google规范是 大括号和 控制语句保持同一行的,我个人也很认
就酱,以后我还会陆续分享,关于代码,求职,学习工作之类的内容。
-------------------------
-* 微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
-* B站:[代码随想录](https://space.bilibili.com/525438321)
+
+
+-----------------------
+* 作者微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
+* B站视频:[代码随想录](https://space.bilibili.com/525438321)
* 知识星球:[代码随想录](https://mp.weixin.qq.com/s/QVF6upVMSbgvZy8lHZS3CQ)
-
-
+
diff --git a/problems/前序/关于时间复杂度,你不知道的都在这里!.md b/problems/前序/关于时间复杂度,你不知道的都在这里!.md
index 3f5bc156..4ca9eede 100644
--- a/problems/前序/关于时间复杂度,你不知道的都在这里!.md
+++ b/problems/前序/关于时间复杂度,你不知道的都在这里!.md
@@ -1,12 +1,12 @@
-
-
+
-
+
+欢迎大家参与本项目,贡献其他语言版本的代码,拥抱开源,让更多学习算法的小伙伴们收益!
+
---------------------------
Carl大胆断言:这可能是你见过对时间复杂度分析最通透的一篇文章了。
@@ -165,10 +165,11 @@ O(2 * n^2 + 10 * n + 1000) < O(3 * n^2),所以说最后省略掉常数项系
-------------------------
-* 微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
-* B站:[代码随想录](https://space.bilibili.com/525438321)
+
+
+-----------------------
+* 作者微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
+* B站视频:[代码随想录](https://space.bilibili.com/525438321)
* 知识星球:[代码随想录](https://mp.weixin.qq.com/s/QVF6upVMSbgvZy8lHZS3CQ)
-
-
+
diff --git a/problems/前序/关于空间复杂度,可能有几个疑问?.md b/problems/前序/关于空间复杂度,可能有几个疑问?.md
index 63940116..0208aa91 100644
--- a/problems/前序/关于空间复杂度,可能有几个疑问?.md
+++ b/problems/前序/关于空间复杂度,可能有几个疑问?.md
@@ -1,11 +1,12 @@
-
-
-
+
-
+
+欢迎大家参与本项目,贡献其他语言版本的代码,拥抱开源,让更多学习算法的小伙伴们收益!
+
+
# 空间复杂度分析
@@ -68,10 +69,11 @@ for (int i = 0; i < n; i++) {
至于如何求递归的空间复杂度,我会在专门写一篇文章来介绍的,敬请期待!
-------------------------
-* 微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
-* B站:[代码随想录](https://space.bilibili.com/525438321)
+
+
+-----------------------
+* 作者微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
+* B站视频:[代码随想录](https://space.bilibili.com/525438321)
* 知识星球:[代码随想录](https://mp.weixin.qq.com/s/QVF6upVMSbgvZy8lHZS3CQ)
-
-
+
diff --git a/problems/前序/刷了这么多题,你了解自己代码的内存消耗么?.md b/problems/前序/刷了这么多题,你了解自己代码的内存消耗么?.md
index df555c3b..6d3b4931 100644
--- a/problems/前序/刷了这么多题,你了解自己代码的内存消耗么?.md
+++ b/problems/前序/刷了这么多题,你了解自己代码的内存消耗么?.md
@@ -1,12 +1,13 @@
-
-
-
-
+
-
+
+欢迎大家参与本项目,贡献其他语言版本的代码,拥抱开源,让更多学习算法的小伙伴们收益!
+
+
+
理解代码的内存消耗,最关键是要知道自己所用编程语言的内存管理。
@@ -145,10 +146,11 @@ char型的数据和int型的数据挨在一起,该int数据从地址1开始,
之后也可以有意识的去学习自己所用的编程语言是如何管理内存的,这些也是程序员的内功。
-------------------------
-* 微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
-* B站:[代码随想录](https://space.bilibili.com/525438321)
+
+
+-----------------------
+* 作者微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
+* B站视频:[代码随想录](https://space.bilibili.com/525438321)
* 知识星球:[代码随想录](https://mp.weixin.qq.com/s/QVF6upVMSbgvZy8lHZS3CQ)
-
-
+
diff --git a/problems/前序/力扣上的代码想在本地编译运行?.md b/problems/前序/力扣上的代码想在本地编译运行?.md
index 04e19021..f4a9b0f3 100644
--- a/problems/前序/力扣上的代码想在本地编译运行?.md
+++ b/problems/前序/力扣上的代码想在本地编译运行?.md
@@ -1,10 +1,11 @@
-
-
+
-
+
+欢迎大家参与本项目,贡献其他语言版本的代码,拥抱开源,让更多学习算法的小伙伴们收益!
+
很多录友都问过我一个问题,就是力扣上的代码如何在本地编译运行?
@@ -62,10 +63,11 @@ int main() {
-------------------------
-* 微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
-* B站:[代码随想录](https://space.bilibili.com/525438321)
+
+
+-----------------------
+* 作者微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
+* B站视频:[代码随想录](https://space.bilibili.com/525438321)
* 知识星球:[代码随想录](https://mp.weixin.qq.com/s/QVF6upVMSbgvZy8lHZS3CQ)
-
-
+
diff --git a/problems/前序/北京互联网公司总结.md b/problems/前序/北京互联网公司总结.md
index f1c9ad9a..4dcaa691 100644
--- a/problems/前序/北京互联网公司总结.md
+++ b/problems/前序/北京互联网公司总结.md
@@ -1,11 +1,12 @@
-
-
-
+
-
+
+欢迎大家参与本项目,贡献其他语言版本的代码,拥抱开源,让更多学习算法的小伙伴们收益!
+
+
# 北京互联网公司总结
@@ -110,11 +111,12 @@
就酱,我也会陆续整理其他城市的互联网公司,希望对大家有所帮助。
-------------------------
-* 微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
-* B站:[代码随想录](https://space.bilibili.com/525438321)
+
+
+
+-----------------------
+* 作者微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
+* B站视频:[代码随想录](https://space.bilibili.com/525438321)
* 知识星球:[代码随想录](https://mp.weixin.qq.com/s/QVF6upVMSbgvZy8lHZS3CQ)
-
-
-
+
diff --git a/problems/前序/广州互联网公司总结.md b/problems/前序/广州互联网公司总结.md
index 47b7bf77..e9b2af00 100644
--- a/problems/前序/广州互联网公司总结.md
+++ b/problems/前序/广州互联网公司总结.md
@@ -1,10 +1,11 @@
-
-
+
-
+
+欢迎大家参与本项目,贡献其他语言版本的代码,拥抱开源,让更多学习算法的小伙伴们收益!
+
# 广州互联网公司总结
@@ -73,11 +74,12 @@
-------------------------
-* 微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
-* B站:[代码随想录](https://space.bilibili.com/525438321)
+
+
+
+-----------------------
+* 作者微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
+* B站视频:[代码随想录](https://space.bilibili.com/525438321)
* 知识星球:[代码随想录](https://mp.weixin.qq.com/s/QVF6upVMSbgvZy8lHZS3CQ)
-
-
-
+
diff --git a/problems/前序/成都互联网公司总结.md b/problems/前序/成都互联网公司总结.md
index 2f32849e..2435ccb2 100644
--- a/problems/前序/成都互联网公司总结.md
+++ b/problems/前序/成都互联网公司总结.md
@@ -1,11 +1,12 @@
-
-
-
+
-
+
+欢迎大家参与本项目,贡献其他语言版本的代码,拥抱开源,让更多学习算法的小伙伴们收益!
+
+
# 成都互联网公司总结
@@ -71,11 +72,12 @@
-------------------------
-* 微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
-* B站:[代码随想录](https://space.bilibili.com/525438321)
+
+
+
+-----------------------
+* 作者微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
+* B站视频:[代码随想录](https://space.bilibili.com/525438321)
* 知识星球:[代码随想录](https://mp.weixin.qq.com/s/QVF6upVMSbgvZy8lHZS3CQ)
-
-
-
+
diff --git a/problems/前序/杭州互联网公司总结.md b/problems/前序/杭州互联网公司总结.md
index 23cd4183..e2691469 100644
--- a/problems/前序/杭州互联网公司总结.md
+++ b/problems/前序/杭州互联网公司总结.md
@@ -1,11 +1,12 @@
-
-
-
+
-
+
+欢迎大家参与本项目,贡献其他语言版本的代码,拥抱开源,让更多学习算法的小伙伴们收益!
+
+
# 杭州互联网公司总结
@@ -82,11 +83,12 @@
-------------------------
-* 微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
-* B站:[代码随想录](https://space.bilibili.com/525438321)
+
+
+
+-----------------------
+* 作者微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
+* B站视频:[代码随想录](https://space.bilibili.com/525438321)
* 知识星球:[代码随想录](https://mp.weixin.qq.com/s/QVF6upVMSbgvZy8lHZS3CQ)
-
-
-
+
diff --git a/problems/前序/深圳互联网公司总结.md b/problems/前序/深圳互联网公司总结.md
index b7d15686..4b68dad6 100644
--- a/problems/前序/深圳互联网公司总结.md
+++ b/problems/前序/深圳互联网公司总结.md
@@ -1,10 +1,11 @@
-
-
+
-
+
+欢迎大家参与本项目,贡献其他语言版本的代码,拥抱开源,让更多学习算法的小伙伴们收益!
+
# 深圳互联网公司总结
@@ -76,11 +77,12 @@
* 广发证券,深交所
* 珍爱网(珍爱网是国内知名的婚恋服务网站之一)
-------------------------
-* 微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
-* B站:[代码随想录](https://space.bilibili.com/525438321)
+
+
+
+-----------------------
+* 作者微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
+* B站视频:[代码随想录](https://space.bilibili.com/525438321)
* 知识星球:[代码随想录](https://mp.weixin.qq.com/s/QVF6upVMSbgvZy8lHZS3CQ)
-
-
-
+
diff --git a/problems/前序/程序员写文档工具.md b/problems/前序/程序员写文档工具.md
index 34f2f777..5530e30f 100644
--- a/problems/前序/程序员写文档工具.md
+++ b/problems/前序/程序员写文档工具.md
@@ -1,12 +1,13 @@
-
-
-
-
+
-
+
+欢迎大家参与本项目,贡献其他语言版本的代码,拥抱开源,让更多学习算法的小伙伴们收益!
+
+
+
# 程序员应该用什么用具来写文档?
@@ -132,10 +133,11 @@ Markdown支持部分html,例如这样
如果还没有掌握markdown的你还在等啥,赶紧使用markdown记录起来吧
-------------------------
-* 微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
-* B站:[代码随想录](https://space.bilibili.com/525438321)
+
+
+-----------------------
+* 作者微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
+* B站视频:[代码随想录](https://space.bilibili.com/525438321)
* 知识星球:[代码随想录](https://mp.weixin.qq.com/s/QVF6upVMSbgvZy8lHZS3CQ)
-
-
+
diff --git a/problems/前序/程序员简历.md b/problems/前序/程序员简历.md
index af52df8e..f9a226df 100644
--- a/problems/前序/程序员简历.md
+++ b/problems/前序/程序员简历.md
@@ -1,12 +1,12 @@
-
-
+
-
+
+欢迎大家参与本项目,贡献其他语言版本的代码,拥抱开源,让更多学习算法的小伙伴们收益!
+
---------------------------
# 程序员的简历应该这么写!!(附简历模板)
@@ -131,9 +131,10 @@ Carl校招社招都拿过大厂的offer,同时也看过很多应聘者的简
就酱,「代码随想录」就是这么干货,Carl多年积累的简历技巧都毫不保留的写出来了,如果感觉对你有帮助,就宣传一波「代码随想录」吧,值得大家的关注!
-------------------------
-* 微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
-* B站:[代码随想录](https://space.bilibili.com/525438321)
+
+-----------------------
+* 作者微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
+* B站视频:[代码随想录](https://space.bilibili.com/525438321)
* 知识星球:[代码随想录](https://mp.weixin.qq.com/s/QVF6upVMSbgvZy8lHZS3CQ)
-
+
diff --git a/problems/前序/递归算法的时间与空间复杂度分析.md b/problems/前序/递归算法的时间与空间复杂度分析.md
index c8ef4723..f1501e8a 100644
--- a/problems/前序/递归算法的时间与空间复杂度分析.md
+++ b/problems/前序/递归算法的时间与空间复杂度分析.md
@@ -1,10 +1,11 @@
-
-
+
-
+
+欢迎大家参与本项目,贡献其他语言版本的代码,拥抱开源,让更多学习算法的小伙伴们收益!
+
# 递归算法的时间与空间复杂度分析!
@@ -263,11 +264,12 @@ int binary_search( int arr[], int l, int r, int x) {
-------------------------
-* 微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
-* B站:[代码随想录](https://space.bilibili.com/525438321)
+
+
+
+-----------------------
+* 作者微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
+* B站视频:[代码随想录](https://space.bilibili.com/525438321)
* 知识星球:[代码随想录](https://mp.weixin.qq.com/s/QVF6upVMSbgvZy8lHZS3CQ)
-
-
-
+
diff --git a/problems/前序/通过一道面试题目,讲一讲递归算法的时间复杂度!.md b/problems/前序/通过一道面试题目,讲一讲递归算法的时间复杂度!.md
index cb6aa604..16ba8361 100644
--- a/problems/前序/通过一道面试题目,讲一讲递归算法的时间复杂度!.md
+++ b/problems/前序/通过一道面试题目,讲一讲递归算法的时间复杂度!.md
@@ -1,11 +1,12 @@
-
-
-
+
-
+
+欢迎大家参与本项目,贡献其他语言版本的代码,拥抱开源,让更多学习算法的小伙伴们收益!
+
+
# 通过一道面试题目,讲一讲递归算法的时间复杂度!
@@ -149,10 +150,11 @@ int function3(int x, int n) {
如果认真读完本篇,相信大家对递归算法的有一个新的认识的,同一道题目,同样是递归,效率可是不一样的!
-------------------------
-* 微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
-* B站:[代码随想录](https://space.bilibili.com/525438321)
+
+
+-----------------------
+* 作者微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
+* B站视频:[代码随想录](https://space.bilibili.com/525438321)
* 知识星球:[代码随想录](https://mp.weixin.qq.com/s/QVF6upVMSbgvZy8lHZS3CQ)
-
-
+
From e5ee38f826a9d230e1f474f1e4130fa564ac7674 Mon Sep 17 00:00:00 2001
From: SkyLazy <627770537@qq.com>
Date: Fri, 4 Jun 2021 14:39:03 +0800
Subject: [PATCH 76/95] =?UTF-8?q?Update=201005.K=E6=AC=A1=E5=8F=96?=
=?UTF-8?q?=E5=8F=8D=E5=90=8E=E6=9C=80=E5=A4=A7=E5=8C=96=E7=9A=84=E6=95=B0?=
=?UTF-8?q?=E7=BB=84=E5=92=8C.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
java 版本长度为1的时候也要考虑k的奇偶性才能直接return吧
---
problems/1005.K次取反后最大化的数组和.md | 2 +-
1 file changed, 1 insertion(+), 1 deletion(-)
diff --git a/problems/1005.K次取反后最大化的数组和.md b/problems/1005.K次取反后最大化的数组和.md
index e57e26ad..5539aff8 100644
--- a/problems/1005.K次取反后最大化的数组和.md
+++ b/problems/1005.K次取反后最大化的数组和.md
@@ -102,7 +102,7 @@ Java:
```java
class Solution {
public int largestSumAfterKNegations(int[] A, int K) {
- if (A.length == 1) return A[0];
+ if (A.length == 1) return k % 2 == 0 ? A[0] : -A[0];
Arrays.sort(A);
int sum = 0;
int idx = 0;
From b69ca00d2fc9a024145f6bc6d88de8ea88e3f3ab Mon Sep 17 00:00:00 2001
From: Baturu <45113401+z80160280@users.noreply.github.com>
Date: Fri, 4 Jun 2021 01:45:10 -0700
Subject: [PATCH 77/95] =?UTF-8?q?Update=200226.=E7=BF=BB=E8=BD=AC=E4=BA=8C?=
=?UTF-8?q?=E5=8F=89=E6=A0=91.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0226.翻转二叉树.md | 50 ++++++++++++++++++++++++++++++++
1 file changed, 50 insertions(+)
diff --git a/problems/0226.翻转二叉树.md b/problems/0226.翻转二叉树.md
index f48eefd2..2b628ec4 100644
--- a/problems/0226.翻转二叉树.md
+++ b/problems/0226.翻转二叉树.md
@@ -230,6 +230,56 @@ class Solution {
Python:
+> 递归法:前序遍历
+```python
+class Solution:
+ def invertTree(self, root: TreeNode) -> TreeNode:
+ if not root:
+ return None
+ root.left, root.right = root.right, root.left #中
+ self.invertTree(root.left) #左
+ self.invertTree(root.right) #右
+ return root
+```
+
+> 迭代法:深度优先遍历(前序遍历)
+```python
+class Solution:
+ def invertTree(self, root: TreeNode) -> TreeNode:
+ if not root:
+ return root
+ st = []
+ st.append(root)
+ while st:
+ node = st.pop()
+ node.left, node.right = node.right, node.left #中
+ if node.right:
+ st.append(node.right) #右
+ if node.left:
+ st.append(node.left) #左
+ return root
+```
+
+> 迭代法:广度优先遍历(层序遍历)
+```python
+import collections
+class Solution:
+ def invertTree(self, root: TreeNode) -> TreeNode:
+ queue = collections.deque() #使用deque()
+ if root:
+ queue.append(root)
+ while queue:
+ size = len(queue)
+ for i in range(size):
+ node = queue.popleft()
+ node.left, node.right = node.right, node.left #节点处理
+ if node.left:
+ queue.append(node.left)
+ if node.right:
+ queue.append(node.right)
+ return root
+```
+
Go:
```Go
func invertTree(root *TreeNode) *TreeNode {
From b655cb2730d1bc60f02d72cfd2918cb4d0789d3f Mon Sep 17 00:00:00 2001
From: Baturu <45113401+z80160280@users.noreply.github.com>
Date: Fri, 4 Jun 2021 02:10:07 -0700
Subject: [PATCH 78/95] =?UTF-8?q?Update=200101.=E5=AF=B9=E7=A7=B0=E4=BA=8C?=
=?UTF-8?q?=E5=8F=89=E6=A0=91.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0101.对称二叉树.md | 72 ++++++++++++++++++++++++++++++++
1 file changed, 72 insertions(+)
diff --git a/problems/0101.对称二叉树.md b/problems/0101.对称二叉树.md
index d8797d30..3dba1648 100644
--- a/problems/0101.对称二叉树.md
+++ b/problems/0101.对称二叉树.md
@@ -360,6 +360,78 @@ Java:
Python:
+> 递归法
+```python
+class Solution:
+ def isSymmetric(self, root: TreeNode) -> bool:
+ if not root:
+ return True
+ return self.compare(root.left, root.right)
+
+ def compare(self, left, right):
+ #首先排除空节点的情况
+ if left == None and right != None: return False
+ elif left != None and right == None: return False
+ elif left == None and right == None: return True
+ #排除了空节点,再排除数值不相同的情况
+ elif left.val != right.val: return False
+
+ #此时就是:左右节点都不为空,且数值相同的情况
+ #此时才做递归,做下一层的判断
+ outside = self.compare(left.left, right.right) #左子树:左、 右子树:右
+ inside = self.compare(left.right, right.left) #左子树:右、 右子树:左
+ isSame = outside and inside #左子树:中、 右子树:中 (逻辑处理)
+ return isSame
+```
+
+> 迭代法: 使用队列
+```python
+import collections
+class Solution:
+ def isSymmetric(self, root: TreeNode) -> bool:
+ if not root:
+ return True
+ queue = collections.deque()
+ queue.append(root.left) #将左子树头结点加入队列
+ queue.append(root.right) #将右子树头结点加入队列
+ while queue: #接下来就要判断这这两个树是否相互翻转
+ leftNode = queue.popleft()
+ rightNode = queue.popleft()
+ if not leftNode and not rightNode: #左节点为空、右节点为空,此时说明是对称的
+ continue
+
+ #左右一个节点不为空,或者都不为空但数值不相同,返回false
+ if not leftNode or not rightNode or leftNode.val != rightNode.val:
+ return False
+ queue.append(leftNode.left) #加入左节点左孩子
+ queue.append(rightNode.right) #加入右节点右孩子
+ queue.append(leftNode.right) #加入左节点右孩子
+ queue.append(rightNode.left) #加入右节点左孩子
+ return True
+```
+
+> 迭代法:使用栈
+```python
+class Solution:
+ def isSymmetric(self, root: TreeNode) -> bool:
+ if not root:
+ return True
+ st = [] #这里改成了栈
+ st.append(root.left)
+ st.append(root.right)
+ while st:
+ leftNode = st.pop()
+ rightNode = st.pop()
+ if not leftNode and not rightNode:
+ continue
+ if not leftNode or not rightNode or leftNode.val != rightNode.val:
+ return False
+ st.append(leftNode.left)
+ st.append(rightNode.right)
+ st.append(leftNode.right)
+ st.append(rightNode.left)
+ return True
+```
Go:
From c44a3165c4333530eac041aa1ede69088ed6462c Mon Sep 17 00:00:00 2001
From: Baturu <45113401+z80160280@users.noreply.github.com>
Date: Fri, 4 Jun 2021 03:08:41 -0700
Subject: [PATCH 79/95] =?UTF-8?q?Update=200104.=E4=BA=8C=E5=8F=89=E6=A0=91?=
=?UTF-8?q?=E7=9A=84=E6=9C=80=E5=A4=A7=E6=B7=B1=E5=BA=A6.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0104.二叉树的最大深度.md | 105 ++++++++++++++++++++++
1 file changed, 105 insertions(+)
diff --git a/problems/0104.二叉树的最大深度.md b/problems/0104.二叉树的最大深度.md
index 756afb68..4bf2b261 100644
--- a/problems/0104.二叉树的最大深度.md
+++ b/problems/0104.二叉树的最大深度.md
@@ -281,6 +281,111 @@ class Solution {
Python:
+104.二叉树的最大深度
+> 递归法:
+```python
+class Solution:
+ def maxDepth(self, root: TreeNode) -> int:
+ return self.getDepth(root)
+
+ def getDepth(self, node):
+ if not node:
+ return 0
+ leftDepth = self.getDepth(node.left) #左
+ rightDepth = self.getDepth(node.right) #右
+ depth = 1 + max(leftDepth, rightDepth) #中
+ return depth
+```
+> 递归法;精简代码
+```python
+class Solution:
+ def maxDepth(self, root: TreeNode) -> int:
+ if not root:
+ return 0
+ return 1 + max(self.maxDepth(root.left), self.maxDepth(root.right))
+```
+
+> 迭代法:
+```python
+import collections
+class Solution:
+ def maxDepth(self, root: TreeNode) -> int:
+ if not root:
+ return 0
+ depth = 0 #记录深度
+ queue = collections.deque()
+ queue.append(root)
+ while queue:
+ size = len(queue)
+ depth += 1
+ for i in range(size):
+ node = queue.popleft()
+ if node.left:
+ queue.append(node.left)
+ if node.right:
+ queue.append(node.right)
+ return depth
+```
+
+559.N叉树的最大深度
+> 递归法:
+```python
+class Solution:
+ def maxDepth(self, root: 'Node') -> int:
+ if not root:
+ return 0
+ depth = 0
+ for i in range(len(root.children)):
+ depth = max(depth, self.maxDepth(root.children[i]))
+ return depth + 1
+```
+
+> 迭代法:
+```python
+import collections
+class Solution:
+ def maxDepth(self, root: 'Node') -> int:
+ queue = collections.deque()
+ if root:
+ queue.append(root)
+ depth = 0 #记录深度
+ while queue:
+ size = len(queue)
+ depth += 1
+ for i in range(size):
+ node = queue.popleft()
+ for j in range(len(node.children)):
+ if node.children[j]:
+ queue.append(node.children[j])
+ return depth
+```
+
+> 使用栈来模拟后序遍历依然可以
+```python
+class Solution:
+ def maxDepth(self, root: 'Node') -> int:
+ st = []
+ if root:
+ st.append(root)
+ depth = 0
+ result = 0
+ while st:
+ node = st.pop()
+ if node != None:
+ st.append(node) #中
+ st.append(None)
+ depth += 1
+ for i in range(len(node.children)): #处理孩子
+ if node.children[i]:
+ st.append(node.children[i])
+
+ else:
+ node = st.pop()
+ depth -= 1
+ result = max(result, depth)
+ return result
+```
+
Go:
From 7bf08e13586b38a14879d1c10668b29ee3145318 Mon Sep 17 00:00:00 2001
From: borninfreedom
Date: Sat, 5 Jun 2021 15:28:13 +0800
Subject: [PATCH 80/95] add python codes of binary tree's level order
---
problems/0102.二叉树的层序遍历.md | 34 +++++++++++++++++++++++
1 file changed, 34 insertions(+)
diff --git a/problems/0102.二叉树的层序遍历.md b/problems/0102.二叉树的层序遍历.md
index 51bd8510..bfb78a43 100644
--- a/problems/0102.二叉树的层序遍历.md
+++ b/problems/0102.二叉树的层序遍历.md
@@ -80,6 +80,40 @@ public:
}
};
```
+python代码:
+
+```python
+# Definition for a binary tree node.
+# class TreeNode:
+# def __init__(self, val=0, left=None, right=None):
+# self.val = val
+# self.left = left
+# self.right = right
+class Solution:
+ def levelOrder(self, root: TreeNode) -> List[List[int]]:
+ if not root:
+ return []
+
+ quene = [root]
+ out_list = []
+
+ while quene:
+ in_list = []
+ for i in range(len(quene)):
+ node = quene.pop(0)
+ in_list.append(node.val)
+ if node.left:
+ quene.append(node.left)
+ if node.right:
+ quene.append(node.right)
+
+ out_list.append(in_list)
+
+ return out_list
+```
+
+
+
javascript代码:
```javascript
From 0ebd5f1a812f7d58d52b440f6b2721b8c8db95d7 Mon Sep 17 00:00:00 2001
From: "qingyi.liu"
Date: Sat, 5 Jun 2021 18:32:40 +0800
Subject: [PATCH 81/95] =?UTF-8?q?=E6=B7=BB=E5=8A=A077.=20=E7=BB=84?=
=?UTF-8?q?=E5=90=88=E4=BC=98=E5=8C=96JavaScript=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0077.组合优化.md | 24 +++++++++++++++++++++++-
1 file changed, 23 insertions(+), 1 deletion(-)
diff --git a/problems/0077.组合优化.md b/problems/0077.组合优化.md
index 31acf4a5..d3e82f09 100644
--- a/problems/0077.组合优化.md
+++ b/problems/0077.组合优化.md
@@ -147,7 +147,7 @@ public:
Java:
-```
+```java
class Solution {
List> result = new ArrayList<>();
LinkedList path = new LinkedList<>();
@@ -220,6 +220,28 @@ func backtrack(n,k,start int,track []int){
}
```
+javaScript:
+
+```js
+var combine = function(n, k) {
+ const res = [], path = [];
+ backtracking(n, k, 1);
+ return res;
+ function backtracking (n, k, i){
+ const len = path.length;
+ if(len === k) {
+ res.push(Array.from(path));
+ return;
+ }
+ for(let a = i; a <= n + len - k + 1; a++) {
+ path.push(a);
+ backtracking(n, k, a + 1);
+ path.pop();
+ }
+ }
+};
+```
+
From faaa67c993869deac75fd0c36ba59f397aad582b Mon Sep 17 00:00:00 2001
From: "qingyi.liu"
Date: Sat, 5 Jun 2021 18:37:44 +0800
Subject: [PATCH 82/95] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=C2=B7216.=E7=BB=84?=
=?UTF-8?q?=E5=90=88=E6=80=BB=E5=92=8CIIIJavaScript=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0216.组合总和III.md | 42 ++++++++++++++++++++++++++++++--
1 file changed, 40 insertions(+), 2 deletions(-)
diff --git a/problems/0216.组合总和III.md b/problems/0216.组合总和III.md
index 0aef7aec..9f75b23d 100644
--- a/problems/0216.组合总和III.md
+++ b/problems/0216.组合总和III.md
@@ -180,7 +180,7 @@ if (sum > targetSum) { // 剪枝操作
最后C++代码如下:
-```
+```c++
class Solution {
private:
vector> result; // 存放结果集
@@ -262,7 +262,7 @@ class Solution {
```
Python:
-```python3
+```py
class Solution:
def combinationSum3(self, k: int, n: int) -> List[List[int]]:
res = [] #存放结果集
@@ -284,6 +284,44 @@ class Solution:
Go:
+javaScript:
+
+```js
+// 等差数列
+var maxV = k => k * (9 + 10 - k) / 2;
+var minV = k => k * (1 + k) / 2;
+var combinationSum3 = function(k, n) {
+ if (k > 9 || k < 1) return [];
+ // if (n > maxV(k) || n < minV(k)) return [];
+ // if (n === maxV(k)) return [Array.from({length: k}).map((v, i) => 9 - i)];
+ // if (n === minV(k)) return [Array.from({length: k}).map((v, i) => i + 1)];
+
+ const res = [], path = [];
+ backtracking(k, n, 1, 0);
+ return res;
+ function backtracking(k, n, i, sum){
+ const len = path.length;
+ if (len > k || sum > n) return;
+ if (maxV(k - len) < n - sum) return;
+ if (minV(k - len) > n - sum) return;
+
+ if(len === k && sum == n) {
+ res.push(Array.from(path));
+ return;
+ }
+
+ const min = Math.min(n - sum, 9 + len - k + 1);
+
+ for(let a = i; a <= min; a++) {
+ path.push(a);
+ sum += a;
+ backtracking(k, n, a + 1, sum);
+ path.pop();
+ sum -= a;
+ }
+ }
+};
+```
From 7cbf0b1fff9675d507a686f4de7c08ef0c3d1394 Mon Sep 17 00:00:00 2001
From: "qingyi.liu"
Date: Sat, 5 Jun 2021 18:40:45 +0800
Subject: [PATCH 83/95] =?UTF-8?q?=E6=B7=BB=E5=8A=A017.=E7=94=B5=E8=AF=9D?=
=?UTF-8?q?=E5=8F=B7=E7=A0=81=E7=9A=84=E5=AD=97=E6=AF=8D=E7=BB=84=E5=90=88?=
=?UTF-8?q?JavaScript=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0017.电话号码的字母组合.md | 33 ++++++++++++++++++--
1 file changed, 30 insertions(+), 3 deletions(-)
diff --git a/problems/0017.电话号码的字母组合.md b/problems/0017.电话号码的字母组合.md
index fa135d8d..aefee698 100644
--- a/problems/0017.电话号码的字母组合.md
+++ b/problems/0017.电话号码的字母组合.md
@@ -137,7 +137,7 @@ for (int i = 0; i < letters.size(); i++) {
关键地方都讲完了,按照[关于回溯算法,你该了解这些!](https://mp.weixin.qq.com/s/gjSgJbNbd1eAA5WkA-HeWw)中的回溯法模板,不难写出如下C++代码:
-```
+```c++
// 版本一
class Solution {
private:
@@ -183,7 +183,7 @@ public:
一些写法,是把回溯的过程放在递归函数里了,例如如下代码,我可以写成这样:(注意注释中不一样的地方)
-```
+```c++
// 版本二
class Solution {
private:
@@ -319,7 +319,7 @@ class Solution:
python3:
-```python3
+```py
class Solution:
def letterCombinations(self, digits: str) -> List[str]:
self.s = ""
@@ -342,6 +342,33 @@ class Solution:
Go:
+javaScript:
+
+```js
+var letterCombinations = function(digits) {
+ const k = digits.length;
+ const map = ["","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"];
+ if(!k) return [];
+ if(k === 1) return map[digits].split("");
+
+ const res = [], path = [];
+ backtracking(digits, k, 0);
+ return res;
+
+ function backtracking(n, k, a) {
+ if(path.length === k) {
+ res.push(path.join(""));
+ return;
+ }
+ for(const v of map[n[a]]) {
+ path.push(v);
+ backtracking(n, k, a + 1);
+ path.pop();
+ }
+
+ }
+};
+```
From d6955ebf03d534045271b74b879a937cdc32e0e0 Mon Sep 17 00:00:00 2001
From: "qingyi.liu"
Date: Sat, 5 Jun 2021 18:42:38 +0800
Subject: [PATCH 84/95] =?UTF-8?q?=E6=B7=BB=E5=8A=A039.=20=E7=BB=84?=
=?UTF-8?q?=E5=90=88=E6=80=BB=E5=92=8CJavaScript=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0039.组合总和.md | 44 +++++++++++++++++------------------
1 file changed, 22 insertions(+), 22 deletions(-)
diff --git a/problems/0039.组合总和.md b/problems/0039.组合总和.md
index e1023ec9..e3e4a117 100644
--- a/problems/0039.组合总和.md
+++ b/problems/0039.组合总和.md
@@ -286,30 +286,30 @@ class Solution:
```
Go:
-JavaScript
+JavaScript:
+
```js
-var strStr = function (haystack, needle) {
- if (needle === '') {
- return 0;
- }
-
- let hayslen = haystack.length;
- let needlen = needle.length;
-
- if (haystack === '' || hayslen < needlen) {
- return -1;
- }
-
- for (let i = 0; i <= hayslen - needlen; i++) {
- if (haystack[i] === needle[0]) {
- if (haystack.substr(i, needlen) === needle) {
- return i;
- }
+var combinationSum = function(candidates, target) {
+ const res = [], path = [];
+ candidates.sort(); // 排序
+ backtracking(0, 0);
+ return res;
+ function backtracking(j, sum) {
+ if (sum > target) return;
+ if (sum === target) {
+ res.push(Array.from(path));
+ return;
+ }
+ for(let i = j; i < candidates.length; i++ ) {
+ const n = candidates[i];
+ if(n > target - sum) continue;
+ path.push(n);
+ sum += n;
+ backtracking(i, sum);
+ path.pop();
+ sum -= n;
+ }
}
- if (i === hayslen - needlen) {
- return -1;
- }
- }
};
```
From ece55fe7efe84e4a76e1fd35d3046dd1aa67f2b5 Mon Sep 17 00:00:00 2001
From: "qingyi.liu"
Date: Sat, 5 Jun 2021 18:44:08 +0800
Subject: [PATCH 85/95] =?UTF-8?q?=E6=B7=BB=E5=8A=A040.=E7=BB=84=E5=90=88?=
=?UTF-8?q?=E6=80=BB=E5=92=8CIIJavaScript=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0040.组合总和II.md | 34 ++++++++++++++++++++++++++++++++-
1 file changed, 33 insertions(+), 1 deletion(-)
diff --git a/problems/0040.组合总和II.md b/problems/0040.组合总和II.md
index 4d2996f5..a462b524 100644
--- a/problems/0040.组合总和II.md
+++ b/problems/0040.组合总和II.md
@@ -293,7 +293,7 @@ class Solution {
}
```
Python:
-```python3
+```py
class Solution:
def combinationSum2(self, candidates: List[int], target: int) -> List[List[int]]:
res = []
@@ -314,7 +314,39 @@ class Solution:
```
Go:
+javaScript:
+```js
+/**
+ * @param {number[]} candidates
+ * @param {number} target
+ * @return {number[][]}
+ */
+var combinationSum2 = function(candidates, target) {
+ const res = []; path = [], len = candidates.length;
+ candidates.sort();
+ backtracking(0, 0);
+ return res;
+ function backtracking(sum, i) {
+ if (sum > target) return;
+ if (sum === target) {
+ res.push(Array.from(path));
+ return;
+ }
+ let f = -1;
+ for(let j = i; j < len; j++) {
+ const n = candidates[j];
+ if(n > target - sum || n === f) continue;
+ path.push(n);
+ sum += n;
+ f = n;
+ backtracking(sum, j + 1);
+ path.pop();
+ sum -= n;
+ }
+ }
+};
+```
-----------------------
From b6b04cc035fa5b82d1a40cb9994f7cdb463bc466 Mon Sep 17 00:00:00 2001
From: "qingyi.liu"
Date: Sat, 5 Jun 2021 18:45:22 +0800
Subject: [PATCH 86/95] =?UTF-8?q?=E6=B7=BB=E5=8A=A0131.=E5=88=86=E5=89=B2?=
=?UTF-8?q?=E5=9B=9E=E6=96=87=E4=B8=B2JavaScript=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0131.分割回文串.md | 33 +++++++++++++++++++++++++++++++-
1 file changed, 32 insertions(+), 1 deletion(-)
diff --git a/problems/0131.分割回文串.md b/problems/0131.分割回文串.md
index c722af37..9d23fd13 100644
--- a/problems/0131.分割回文串.md
+++ b/problems/0131.分割回文串.md
@@ -292,7 +292,7 @@ class Solution {
```
Python:
-```python3
+```py
class Solution:
def partition(self, s: str) -> List[List[str]]:
res = []
@@ -313,7 +313,38 @@ class Solution:
Go:
+javaScript:
+```js
+/**
+ * @param {string} s
+ * @return {string[][]}
+ */
+const isPalindrome = (s, l, r) => {
+ for (let i = l, j = r; i < j; i++, j--) {
+ if(s[i] !== s[j]) return false;
+ }
+ return true;
+}
+
+var partition = function(s) {
+ const res = [], path = [], len = s.length;
+ backtracking(0);
+ return res;
+ function backtracking(i) {
+ if(i >= len) {
+ res.push(Array.from(path));
+ return;
+ }
+ for(let j = i; j < len; j++) {
+ if(!isPalindrome(s, i, j)) continue;
+ path.push(s.substr(i, j - i + 1));
+ backtracking(j + 1);
+ path.pop();
+ }
+ }
+};
+```
-----------------------
From 533964af4b31bf1d08f45e626f562616ae12ca8b Mon Sep 17 00:00:00 2001
From: "qingyi.liu"
Date: Sat, 5 Jun 2021 18:46:22 +0800
Subject: [PATCH 87/95] =?UTF-8?q?=E6=B7=BB=E5=8A=A093.=E5=A4=8D=E5=8E=9FIP?=
=?UTF-8?q?=E5=9C=B0=E5=9D=80JavaScript=C2=B7=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0093.复原IP地址.md | 29 +++++++++++++++++++++++++++++
1 file changed, 29 insertions(+)
diff --git a/problems/0093.复原IP地址.md b/problems/0093.复原IP地址.md
index 3c8e5d9d..a8b9a215 100644
--- a/problems/0093.复原IP地址.md
+++ b/problems/0093.复原IP地址.md
@@ -338,6 +338,35 @@ class Solution(object):
return ans```
```
+JavaScript:
+
+```js
+/**
+ * @param {string} s
+ * @return {string[]}
+ */
+var restoreIpAddresses = function(s) {
+ const res = [], path = [];
+ backtracking(0, 0)
+ return res;
+ function backtracking(i) {
+ const len = path.length;
+ if(len > 4) return;
+ if(len === 4 && i === s.length) {
+ res.push(path.join("."));
+ return;
+ }
+ for(let j = i; j < s.length; j++) {
+ const str = s.substr(i, j - i + 1);
+ if(str.length > 3 || +str > 255) break;
+ if(str.length > 1 && str[0] === "0") break;
+ path.push(str);
+ backtracking(j + 1);
+ path.pop()
+ }
+ }
+};
+```
-----------------------
* 作者微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
From 03ff3ad0cb4359705ffc5ff08059c83c95fbda73 Mon Sep 17 00:00:00 2001
From: borninfreedom
Date: Sat, 5 Jun 2021 19:07:12 +0800
Subject: [PATCH 88/95] =?UTF-8?q?=E7=BB=99=E5=85=B6=E4=BB=96=E5=87=A0?=
=?UTF-8?q?=E9=81=93=E9=A2=98=E7=9B=AE=E6=B7=BB=E5=8A=A0=E4=BA=86python?=
=?UTF-8?q?=E7=9A=84=E4=BB=A3=E7=A0=81?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0102.二叉树的层序遍历.md | 184 +++++++++++++++++++++-
1 file changed, 183 insertions(+), 1 deletion(-)
diff --git a/problems/0102.二叉树的层序遍历.md b/problems/0102.二叉树的层序遍历.md
index bfb78a43..c7fdf776 100644
--- a/problems/0102.二叉树的层序遍历.md
+++ b/problems/0102.二叉树的层序遍历.md
@@ -99,7 +99,7 @@ class Solution:
while quene:
in_list = []
- for i in range(len(quene)):
+ for _ in range(len(quene)):
node = quene.pop(0)
in_list.append(node.val)
if node.left:
@@ -185,6 +185,43 @@ public:
}
};
```
+python代码:
+
+```python
+# Definition for a binary tree node.
+# class TreeNode:
+# def __init__(self, val=0, left=None, right=None):
+# self.val = val
+# self.left = left
+# self.right = right
+class Solution:
+ def levelOrderBottom(self, root: TreeNode) -> List[List[int]]:
+ if not root:
+ return []
+ quene = [root]
+ out_list = []
+
+ while quene:
+ in_list = []
+ for _ in range(len(quene)):
+ node = quene.pop(0)
+ in_list.append(node.val)
+ if node.left:
+ quene.append(node.left)
+ if node.right:
+ quene.append(node.right)
+
+ out_list.append(in_list)
+
+ out_list.reverse()
+ return out_list
+
+# 执行用时:36 ms, 在所有 Python3 提交中击败了92.00%的用户
+# 内存消耗:15.2 MB, 在所有 Python3 提交中击败了63.76%的用户
+```
+
+
+
javascript代码
```javascript
@@ -246,6 +283,50 @@ public:
}
};
```
+python代码:
+
+```python
+# Definition for a binary tree node.
+# class TreeNode:
+# def __init__(self, val=0, left=None, right=None):
+# self.val = val
+# self.left = left
+# self.right = right
+class Solution:
+ def rightSideView(self, root: TreeNode) -> List[int]:
+ if not root:
+ return []
+
+ # deque来自collections模块,不在力扣平台时,需要手动写入
+ # 'from collections import deque' 导入
+ # deque相比list的好处是,list的pop(0)是O(n)复杂度,deque的popleft()是O(1)复杂度
+
+ quene = deque([root])
+ out_list = []
+
+ while quene:
+ # 每次都取最后一个node就可以了
+ node = quene[-1]
+ out_list.append(node.val)
+
+ # 执行这个遍历的目的是获取下一层所有的node
+ for _ in range(len(quene)):
+ node = quene.popleft()
+ if node.left:
+ quene.append(node.left)
+ if node.right:
+ quene.append(node.right)
+
+ return out_list
+
+# 执行用时:36 ms, 在所有 Python3 提交中击败了89.47%的用户
+# 内存消耗:14.6 MB, 在所有 Python3 提交中击败了96.65%的用户
+```
+
+
+
+
+
javascript代码:
```javascript
@@ -309,6 +390,46 @@ public:
```
+python代码:
+
+```python
+# Definition for a binary tree node.
+# class TreeNode:
+# def __init__(self, val=0, left=None, right=None):
+# self.val = val
+# self.left = left
+# self.right = right
+class Solution:
+ def averageOfLevels(self, root: TreeNode) -> List[float]:
+ if not root:
+ return []
+
+ quene = deque([root])
+ out_list = []
+
+ while quene:
+ in_list = []
+
+ for _ in range(len(quene)):
+ node = quene.popleft()
+ in_list.append(node.val)
+ if node.left:
+ quene.append(node.left)
+ if node.right:
+ quene.append(node.right)
+
+ out_list.append(in_list)
+
+ out_list = map(lambda x: sum(x) / len(x), out_list)
+
+ return out_list
+
+# 执行用时:56 ms, 在所有 Python3 提交中击败了81.48%的用户
+# 内存消耗:17 MB, 在所有 Python3 提交中击败了89.68%的用户
+```
+
+
+
javascript代码:
```javascript
@@ -385,7 +506,68 @@ public:
};
```
+python代码:
+
+```python
+"""
+# Definition for a Node.
+class Node:
+ def __init__(self, val=None, children=None):
+ self.val = val
+ self.children = children
+"""
+
+class Solution:
+ def levelOrder(self, root: 'Node') -> List[List[int]]:
+ if not root:
+ return []
+
+ quene = deque([root])
+ out_list = []
+
+ while quene:
+ in_list = []
+
+ for _ in range(len(quene)):
+ node = quene.popleft()
+ in_list.append(node.val)
+ if node.children:
+ # 这个地方要用extend而不是append,我们看下面的例子:
+ # In [18]: alist=[]
+ # In [19]: alist.append([1,2,3])
+ # In [20]: alist
+ # Out[20]: [[1, 2, 3]]
+ # In [21]: alist.extend([4,5,6])
+ # In [22]: alist
+ # Out[22]: [[1, 2, 3], 4, 5, 6]
+ # 可以看到extend对要添加的list进行了一个解包操作
+ # print(root.children),可以得到children是一个包含
+ # 孩子节点地址的list,我们使用for遍历quene的时候,
+ # 希望quene是一个单层list,所以要用extend
+ # 使用extend的情况,如果print(quene),结果是
+ # deque([<__main__.Node object at 0x7f60763ae0a0>])
+ # deque([<__main__.Node object at 0x7f607636e6d0>, <__main__.Node object at 0x7f607636e130>, <__main__.Node object at 0x7f607636e310>])
+ # deque([<__main__.Node object at 0x7f607636e880>, <__main__.Node object at 0x7f607636ef10>])
+ # 可以看到是单层list
+ # 如果使用append,print(quene)的结果是
+ # deque([<__main__.Node object at 0x7f18907530a0>])
+ # deque([[<__main__.Node object at 0x7f18907136d0>, <__main__.Node object at 0x7f1890713130>, <__main__.Node object at 0x7f1890713310>]])
+ # 可以看到是两层list,这样for的遍历就会报错
+
+ quene.extend(node.children)
+
+ out_list.append(in_list)
+
+ return out_list
+
+# 执行用时:60 ms, 在所有 Python3 提交中击败了76.99%的用户
+# 内存消耗:16.5 MB, 在所有 Python3 提交中击败了89.19%的用户
+```
+
+
+
JavaScript代码:
+
```JavaScript
var levelOrder = function(root) {
//每一层可能有2个以上,所以不再使用node.left node.right
From 546865ce4c18c00589dad748e9f8c7eaa29ede80 Mon Sep 17 00:00:00 2001
From: LiangDazhu <42199191+LiangDazhu@users.noreply.github.com>
Date: Sun, 6 Jun 2021 00:44:21 +0800
Subject: [PATCH 89/95] =?UTF-8?q?Update=200474.=E4=B8=80=E5=92=8C=E9=9B=B6?=
=?UTF-8?q?.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
Added python version code
---
problems/0474.一和零.md | 13 ++++++++++++-
1 file changed, 12 insertions(+), 1 deletion(-)
diff --git a/problems/0474.一和零.md b/problems/0474.一和零.md
index e158ca63..a098b9ef 100644
--- a/problems/0474.一和零.md
+++ b/problems/0474.一和零.md
@@ -190,7 +190,18 @@ class Solution {
```
Python:
-
+```python
+class Solution:
+ def findMaxForm(self, strs: List[str], m: int, n: int) -> int:
+ dp = [[0] * (n + 1) for _ in range(m + 1)]
+ for str in strs:
+ oneNum = str.count('1')
+ zeroNum = str.count('0')
+ for i in range(m, zeroNum - 1, -1):
+ for j in range(n, oneNum - 1, -1):
+ dp[i][j] = max(dp[i][j], dp[i - zeroNum][j - oneNum] + 1)
+ return dp[m][n]
+```
Go:
From d25bdef4a515064542c064abc073f09e29ce104c Mon Sep 17 00:00:00 2001
From: Yang
Date: Sat, 5 Jun 2021 20:14:18 -0400
Subject: [PATCH 90/95] =?UTF-8?q?Update=200377.=E7=BB=84=E5=90=88=E6=80=BB?=
=?UTF-8?q?=E5=92=8C=E2=85=A3.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
Separate Java and Python code
---
problems/0377.组合总和Ⅳ.md | 2 +-
1 file changed, 1 insertion(+), 1 deletion(-)
diff --git a/problems/0377.组合总和Ⅳ.md b/problems/0377.组合总和Ⅳ.md
index 6813f13c..c6dc3d42 100644
--- a/problems/0377.组合总和Ⅳ.md
+++ b/problems/0377.组合总和Ⅳ.md
@@ -163,7 +163,7 @@ class Solution {
return dp[target];
}
}
-
+```
Python:
From ea6dcc9d8f5962e13bdd9f8034f244bfec450c42 Mon Sep 17 00:00:00 2001
From: Baturu <45113401+z80160280@users.noreply.github.com>
Date: Sat, 5 Jun 2021 17:33:44 -0700
Subject: [PATCH 91/95] =?UTF-8?q?Update=200222.=E5=AE=8C=E5=85=A8=E4=BA=8C?=
=?UTF-8?q?=E5=8F=89=E6=A0=91=E7=9A=84=E8=8A=82=E7=82=B9=E4=B8=AA=E6=95=B0?=
=?UTF-8?q?.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
.../0222.完全二叉树的节点个数.md | 65 +++++++++++++++++++
1 file changed, 65 insertions(+)
diff --git a/problems/0222.完全二叉树的节点个数.md b/problems/0222.完全二叉树的节点个数.md
index 2e2fed99..91e24247 100644
--- a/problems/0222.完全二叉树的节点个数.md
+++ b/problems/0222.完全二叉树的节点个数.md
@@ -240,6 +240,71 @@ class Solution {
Python:
+> 递归法:
+```python
+class Solution:
+ def countNodes(self, root: TreeNode) -> int:
+ return self.getNodesNum(root)
+
+ def getNodesNum(self, cur):
+ if not cur:
+ return 0
+ leftNum = self.getNodesNum(cur.left) #左
+ rightNum = self.getNodesNum(cur.right) #右
+ treeNum = leftNum + rightNum + 1 #中
+ return treeNum
+```
+
+> 递归法:精简版
+```python
+class Solution:
+ def countNodes(self, root: TreeNode) -> int:
+ if not root:
+ return 0
+ return 1 + self.countNodes(root.left) + self.countNodes(root.right)
+```
+
+> 迭代法:
+```python
+import collections
+class Solution:
+ def countNodes(self, root: TreeNode) -> int:
+ queue = collections.deque()
+ if root:
+ queue.append(root)
+ result = 0
+ while queue:
+ size = len(queue)
+ for i in range(size):
+ node = queue.popleft()
+ result += 1 #记录节点数量
+ if node.left:
+ queue.append(node.left)
+ if node.right:
+ queue.append(node.right)
+ return result
+```
+
+> 完全二叉树
+```python
+class Solution:
+ def countNodes(self, root: TreeNode) -> int:
+ if not root:
+ return 0
+ left = root.left
+ right = root.right
+ leftHeight = 0 #这里初始为0是有目的的,为了下面求指数方便
+ rightHeight = 0
+ while left: #求左子树深度
+ left = left.left
+ leftHeight += 1
+ while right: #求右子树深度
+ right = right.right
+ rightHeight += 1
+ if leftHeight == rightHeight:
+ return (2 << leftHeight) - 1 #注意(2<<1) 相当于2^2,所以leftHeight初始为0
+ return self.countNodes(root.left) + self.countNodes(root.right) + 1
+```
Go:
From 6efc66e17597da3038966c690bb75dd4f0e61ad0 Mon Sep 17 00:00:00 2001
From: Baturu <45113401+z80160280@users.noreply.github.com>
Date: Sat, 5 Jun 2021 18:13:20 -0700
Subject: [PATCH 92/95] =?UTF-8?q?Update=200110.=E5=B9=B3=E8=A1=A1=E4=BA=8C?=
=?UTF-8?q?=E5=8F=89=E6=A0=91.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0110.平衡二叉树.md | 56 ++++++++++++++++++++++++++++++++
1 file changed, 56 insertions(+)
diff --git a/problems/0110.平衡二叉树.md b/problems/0110.平衡二叉树.md
index 1cd54849..b9d01503 100644
--- a/problems/0110.平衡二叉树.md
+++ b/problems/0110.平衡二叉树.md
@@ -498,6 +498,62 @@ class Solution {
Python:
+> 递归法:
+```python
+class Solution:
+ def isBalanced(self, root: TreeNode) -> bool:
+ return True if self.getDepth(root) != -1 else False
+
+ #返回以该节点为根节点的二叉树的高度,如果不是二叉搜索树了则返回-1
+ def getDepth(self, node):
+ if not node:
+ return 0
+ leftDepth = self.getDepth(node.left)
+ if leftDepth == -1: return -1 #说明左子树已经不是二叉平衡树
+ rightDepth = self.getDepth(node.right)
+ if rightDepth == -1: return -1 #说明右子树已经不是二叉平衡树
+ return -1 if abs(leftDepth - rightDepth)>1 else 1 + max(leftDepth, rightDepth)
+```
+
+> 迭代法:
+```python
+class Solution:
+ def isBalanced(self, root: TreeNode) -> bool:
+ st = []
+ if not root:
+ return True
+ st.append(root)
+ while st:
+ node = st.pop() #中
+ if abs(self.getDepth(node.left) - self.getDepth(node.right)) > 1:
+ return False
+ if node.right:
+ st.append(node.right) #右(空节点不入栈)
+ if node.left:
+ st.append(node.left) #左(空节点不入栈)
+ return True
+
+ def getDepth(self, cur):
+ st = []
+ if cur:
+ st.append(cur)
+ depth = 0
+ result = 0
+ while st:
+ node = st.pop()
+ if node:
+ st.append(node) #中
+ st.append(None)
+ depth += 1
+ if node.right: st.append(node.right) #右
+ if node.left: st.append(node.left) #左
+ else:
+ node = st.pop()
+ depth -= 1
+ result = max(result, depth)
+ return result
+```
+
Go:
```Go
From d68e8922ceea44b9b066025cbbe61e5762d61216 Mon Sep 17 00:00:00 2001
From: Jack Choi <648074155@qq.com>
Date: Sun, 6 Jun 2021 11:05:52 +0800
Subject: [PATCH 93/95] =?UTF-8?q?Update=200027.=E7=A7=BB=E9=99=A4=E5=85=83?=
=?UTF-8?q?=E7=B4=A0.md?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0027.移除元素.md | 18 +++++++++++++++++-
1 file changed, 17 insertions(+), 1 deletion(-)
diff --git a/problems/0027.移除元素.md b/problems/0027.移除元素.md
index f0f61d06..f1187db7 100644
--- a/problems/0027.移除元素.md
+++ b/problems/0027.移除元素.md
@@ -199,7 +199,23 @@ def remove_element(nums, val)
i
end
```
-
+Rust:
+```rust
+pub fn remove_element(nums: &mut Vec, val: i32) -> &mut Vec {
+ let mut start: usize = 0;
+ while start < nums.len() {
+ if nums[start] == val {
+ nums.remove(start);
+ }
+ start += 1;
+ }
+ nums
+}
+fn main() {
+ let mut nums = vec![5,1,3,5,2,3,4,1];
+ println!("{:?}",remove_element(&mut nums, 5));
+}
+```
-----------------------
* 作者微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
* B站视频:[代码随想录](https://space.bilibili.com/525438321)
From 4dda01ba256a0584fbe20a88ceb73335eb27269c Mon Sep 17 00:00:00 2001
From: haofeng <852172305@qq.com>
Date: Sun, 6 Jun 2021 11:44:12 +0800
Subject: [PATCH 94/95] =?UTF-8?q?Update=200474.=E4=B8=80=E5=92=8C=E9=9B=B6?=
=?UTF-8?q?.md=20=E6=B7=BB=E5=8A=A0=E4=B8=80=E5=92=8C=E9=9B=B6=20python3?=
=?UTF-8?q?=20=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0474.一和零.md | 25 +++++++++++++++++++++++++
1 file changed, 25 insertions(+)
diff --git a/problems/0474.一和零.md b/problems/0474.一和零.md
index e158ca63..1e0dc794 100644
--- a/problems/0474.一和零.md
+++ b/problems/0474.一和零.md
@@ -191,6 +191,31 @@ class Solution {
Python:
+```python3
+class Solution:
+ def countBinary(self, s: str) -> (int, int):
+ zeros, ones = 0, 0
+ for c in s:
+ if c == '0':
+ zeros += 1
+ else:
+ ones += 1
+ return zeros, ones
+
+ def findMaxForm(self, strs: List[str], m: int, n: int) -> int:
+ dp = [[0]*(n + 1) for _ in range(m + 1)] # 默认初始化0
+ # 遍历物品
+ for i in range(len(strs)):
+ # num_zeros = strs[i].count('0')
+ # num_ones = strs[i].count('1')
+ num_zeros, num_ones = self.countBinary(strs[i])
+ # 遍历背包容量且从后向前遍历!
+ for j in range(m, num_zeros - 1, -1):
+ for i in range(n, num_ones - 1, -1):
+ dp[j][i] = max(dp[j - num_zeros][i - num_ones] + 1, dp[j][i])
+ return dp[m][n]
+```
+
Go:
From 9c0fe0ff494aa14ba28816b7f5fec6e6a7b61687 Mon Sep 17 00:00:00 2001
From: haofeng <852172305@qq.com>
Date: Sun, 6 Jun 2021 15:55:35 +0800
Subject: [PATCH 95/95] =?UTF-8?q?Update=20=E8=83=8C=E5=8C=85=E9=97=AE?=
=?UTF-8?q?=E9=A2=98=E7=90=86=E8=AE=BA=E5=9F=BA=E7=A1=80=E5=AE=8C=E5=85=A8?=
=?UTF-8?q?=E8=83=8C=E5=8C=85.md=20=E6=B7=BB=E5=8A=A0=20python3=20?=
=?UTF-8?q?=E7=89=88=E6=9C=AC=E4=BB=A3=E7=A0=81?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
.../背包问题理论基础完全背包.md | 38 ++++++++++++++++++-
1 file changed, 37 insertions(+), 1 deletion(-)
diff --git a/problems/背包问题理论基础完全背包.md b/problems/背包问题理论基础完全背包.md
index 02785ad7..9997dff3 100644
--- a/problems/背包问题理论基础完全背包.md
+++ b/problems/背包问题理论基础完全背包.md
@@ -179,9 +179,45 @@ int main() {
Java:
-
Python:
+```python3
+# 先遍历物品,再遍历背包
+def test_complete_pack1():
+ weight = [1, 3, 4]
+ value = [15, 20, 30]
+ bag_weight = 4
+
+ dp = [0]*(bag_weight + 1)
+
+ for i in range(len(weight)):
+ for j in range(weight[i], bag_weight + 1):
+ dp[j] = max(dp[j], dp[j - weight[i]] + value[i])
+
+ print(dp[bag_weight])
+
+# 先遍历背包,再遍历物品
+def test_complete_pack2():
+ weight = [1, 3, 4]
+ value = [15, 20, 30]
+ bag_weight = 4
+
+ dp = [0]*(bag_weight + 1)
+
+ for j in range(bag_weight + 1):
+ for i in range(len(weight)):
+ if j >= weight[i]: dp[j] = max(dp[j], dp[j - weight[i]] + value[i])
+
+ print(dp[bag_weight])
+
+
+if __name__ == '__main__':
+ test_complete_pack1()
+ test_complete_pack2()
+```
+
+
+
Go: