From abb9cfd5be04acf55f41f0438abe43b587354b3e Mon Sep 17 00:00:00 2001 From: liyao <-> Date: Sat, 14 Aug 2021 16:24:20 +0100 Subject: [PATCH 01/35] =?UTF-8?q?=E6=9B=B4=E6=AD=A3=20=E6=9C=80=E9=95=BF?= =?UTF-8?q?=E9=80=92=E5=A2=9E=E5=AD=90=E5=BA=8F=E5=88=97=20=E7=AE=97?= =?UTF-8?q?=E6=B3=95=E5=A4=8D=E6=9D=82=E5=BA=A6=EF=BC=8C=E6=B7=BB=E5=8A=A0?= =?UTF-8?q?O(nlogn)=E6=97=B6=E9=97=B4=E5=A4=8D=E6=9D=82=E5=BA=A6=E7=AE=97?= =?UTF-8?q?=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0300.最长上升子序列.md | 124 ++++++++++++++++++++++++- 1 file changed, 122 insertions(+), 2 deletions(-) diff --git a/problems/0300.最长上升子序列.md b/problems/0300.最长上升子序列.md index 57edd13e..1161e158 100644 --- a/problems/0300.最长上升子序列.md +++ b/problems/0300.最长上升子序列.md @@ -33,7 +33,7 @@ * 1 <= nums.length <= 2500 * -10^4 <= nums[i] <= 104 - +## 方法一 动态规划 ## 思路 最长上升子序列是动规的经典题目，这里dp[i]是可以根据dp[j] （j < i）推导出来的，那么依然用动规五部曲来分析详细一波： @@ -190,10 +190,130 @@ const lengthOfLIS = (nums) => { }; ``` *复杂度分析* -- 时间复杂度：O(nlogn)。数组 nums 的长度为 n，我们依次用数组中的元素去更新 dp 数组，相当于插入最后递增的元素，而更新 dp 数组时需要进行 O(logn) 的二分搜索，所以总时间复杂度为 O(nlogn)。 +- 时间复杂度：O(n^2)。数组 nums 的长度为 n，我们依次用数组中的元素去遍历 dp 数组，而遍历 dp 数组时需要进行 O(n) 次搜索，所以总时间复杂度为 O(n^2)。 - 空间复杂度：O(n)，需要额外使用长度为 n 的 dp 数组。 +## 方法二 贪心策略+二分搜索 + +使用贪心策略和二分搜索可以进一步将算法时间复杂度将为O(nlogn)。 + +## 思路 + +为了使得到的子序列尽可能长，我们需要使序列上升得尽可能慢。 + +对于长度为n的数组 nums，我们从0到n-1依次遍历数组中的每个元素nums[i],更新在0到i范围内最长上升子序列的长度len，以及 在0到i范围内，上升子序列的长度为1到len时，对应长度子序列最右端的最小值，将结果保存在list中。实际编码过程中，list长度即为len。 + +## 可行性 +当我们遍历完数组nums中第n-1个元素时，list中保存的是0到n-1范围内最长上升子序列的长度，即为所求。 + +## 算法复杂度分析 +1. list中的元素是单调递增的。可以用反证法来证明：假设对于0<=i=list[j]，那么我们可以在list[j]对应的子序列中删除最后j-i个元素得到长度与list[i]相同的子序列，其最右端的值maxlist[len-1],此时，list中子序列长度为1到len的对应的最右端最小值不变，并新增长度为len+1的子序列，最右端的最小值为nums[i],时间复杂度O(1); + + 2. if(nums[i]<=list[len-1])，此时，我们可以在0到len-1范围内找到k,list[k]为>=nums[i]的最小值,由于list单调递增，所以我们可以使用二分搜索在O(logn)的时间复杂度内找到k。 + 1. 对于0<=jnums[i]; + +3. 综上，算法时间复杂度为O(nlogn),空间复杂度为O(n),需要O(n)的空间保存list。 + +代码如下 + +Java +```java +class Solution { + public int lengthOfLIS(int[] nums) { + int n = nums.length; + if(n==0){return 0;} + + List list=new ArrayList<>(); + list.add(nums[0]); + for (int i = 1; i < n; ++i) { + if (nums[i] > list.get(list.size()-1)) { + list.add(nums[i]); + } else { + int k=binarySearch(list,nums[i]); + list.set(k,nums[i]); + } + } + return list.size(); + } + + int binarySearch(Listlist, int num){ + int len=list.size(); + int l=0,r=len-1,ans=len-1; + while(l<=r){ + int mid=l+(r-l)/2; + if(list.get(mid) list[len-1]) { + list[len++]=nums[i]; + } else { + int k=binarySearch(list,len,nums[i]); + list[k]=nums[i]; + } + } + return len; + } + + int binarySearch(int[] list,int len, int num){ + + int l=0,r=len-1,ans=len-1; + while(l<=r){ + int mid=l+(r-l)/2; + if(list[mid] Date: Sun, 15 Aug 2021 22:26:57 +0800 Subject: [PATCH 02/35] =?UTF-8?q?=E6=B7=BB=E5=8A=A00042.=E6=8E=A5=E9=9B=A8?= =?UTF-8?q?=E6=B0=B4java=E5=8F=8C=E6=8C=87=E9=92=88=E8=A7=A3=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0042.接雨水.md | 24 ++++++++++++++++++++++++ 1 file changed, 24 insertions(+) diff --git a/problems/0042.接雨水.md b/problems/0042.接雨水.md index 235776a0..b3ee84fd 100644 --- a/problems/0042.接雨水.md +++ b/problems/0042.接雨水.md @@ -365,7 +365,31 @@ public: ## 其他语言版本 Java: +双指针法 +```java +class Solution { + public int trap(int[] height) { + int sum = 0; + for (int i = 0; i < height.length; i++) { + // 第一个柱子和最后一个柱子不接雨水 + if (i==0 || i== height.length - 1) continue; + + int rHeight = height[i]; // 记录右边柱子的最高高度 + int lHeight = height[i]; // 记录左边柱子的最高高度 + for (int r = i+1; r < height.length; r++) { + if (height[r] > rHeight) rHeight = height[r]; + } + for (int l = i-1; l >= 0; l--) { + if(height[l] > lHeight) lHeight = height[l]; + } + int h = Math.min(lHeight, rHeight) - height[i]; + if (h > 0) sum += h; + } + return sum; + } +} +``` Python: 双指针法 From 2405f03d94320d195c97e9719853297d5373aeaf Mon Sep 17 00:00:00 2001 From: ironartisan Date: Tue, 17 Aug 2021 11:21:23 +0800 Subject: [PATCH 03/35] =?UTF-8?q?=E6=B7=BB=E5=8A=A00042.=E6=8E=A5=E9=9B=A8?= =?UTF-8?q?=E6=B0=B4java=E5=8A=A8=E6=80=81=E8=A7=84=E5=88=92=E8=A7=A3?= =?UTF-8?q?=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0042.接雨水.md | 28 ++++++++++++++++++++++++++++ 1 file changed, 28 insertions(+) diff --git a/problems/0042.接雨水.md b/problems/0042.接雨水.md index c241b441..22c8c0ef 100644 --- a/problems/0042.接雨水.md +++ b/problems/0042.接雨水.md @@ -365,6 +365,7 @@ public: ## 其他语言版本 Java: + 双指针法 ```java class Solution { @@ -390,6 +391,33 @@ class Solution { } } ``` +动态规划法 +```java +class Solution { + public int trap(int[] height) { + int length = height.length; + if (length <= 2) return 0; + int[] maxLeft = new int[length]; + int[] maxRight = new int[length]; + + // 记录每个柱子左边柱子最大高度 + maxLeft[0] = height[0]; + for (int i = 1; i< length; i++) maxLeft[i] = Math.max(height[i], maxLeft[i-1]); + + // 记录每个柱子右边柱子最大高度 + maxRight[length - 1] = height[length - 1]; + for(int i = length - 2; i >= 0; i--) maxRight[i] = Math.max(height[i], maxRight[i+1]); + + // 求和 + int sum = 0; + for (int i = 0; i < length; i++) { + int count = Math.min(maxLeft[i], maxRight[i]) - height[i]; + if (count > 0) sum += count; + } + return sum; + } +} +``` Python: 双指针法 From 20182c753d5451616aea5d7f131ba9cb4fd4f79e Mon Sep 17 00:00:00 2001 From: Nixiak-nan <70318059+Nixiak-nan@users.noreply.github.com> Date: Tue, 17 Aug 2021 12:24:54 +0800 Subject: [PATCH 04/35] =?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=85-1.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 修改错别字 --- problems/背包理论基础01背包-1.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/背包理论基础01背包-1.md b/problems/背包理论基础01背包-1.md index c13efc3a..851c2657 100644 --- a/problems/背包理论基础01背包-1.md +++ b/problems/背包理论基础01背包-1.md @@ -103,7 +103,7 @@ 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 那么很明显当 j < weight[0]的时候，dp[0][j] 应该是 0，因为背包容量比编号0的物品重量还小。 -当j >= weight[0]是，dp[0][j] 应该是value[0]，因为背包容量放足够放编号0物品。 +当j >= weight[0]时，dp[0][j] 应该是value[0]，因为背包容量放足够放编号0物品。 代码初始化如下： ``` From 07e95e7a97f1bdf55386cbc6c05f6b9cfba88a04 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E6=9E=81=E5=AE=A2=E5=AD=A6=E4=BC=9F?= Date: Mon, 16 Aug 2021 13:00:02 +0800 Subject: [PATCH 05/35] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=20203.=E7=A7=BB?= =?UTF-8?q?=E9=99=A4=E9=93=BE=E8=A1=A8=E5=85=83=E7=B4=A0=20Swift=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/0203.移除链表元素.md | 3 +-- 1 file changed, 1 insertion(+), 2 deletions(-) diff --git a/problems/0203.移除链表元素.md b/problems/0203.移除链表元素.md index 9235d47e..f3724fc2 100644 --- a/problems/0203.移除链表元素.md +++ b/problems/0203.移除链表元素.md @@ -304,8 +304,7 @@ var removeElements = function(head, val) { }; ``` -Swift: - +Swift： ```swift /** * Definition for singly-linked list. From c7c52f7db6d92800a273057c3a5abc4cffcc8aef Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E6=9E=81=E5=AE=A2=E5=AD=A6=E4=BC=9F?= Date: Mon, 16 Aug 2021 20:28:38 +0800 Subject: [PATCH 06/35] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=20707.=E8=AE=BE?= =?UTF-8?q?=E8=AE=A1=E9=93=BE=E8=A1=A8=20Swift=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0707.设计链表.md | 64 +++++++++++++++++++++++++++++++++++ 1 file changed, 64 insertions(+) diff --git a/problems/0707.设计链表.md b/problems/0707.设计链表.md index 0aa038e8..33b02e56 100644 --- a/problems/0707.设计链表.md +++ b/problems/0707.设计链表.md @@ -948,6 +948,70 @@ class MyLinkedList { } ``` +Swift +```Swift +class MyLinkedList { + var size = 0 + let head: ListNode + + /** Initialize your data structure here. */ + init() { + head = ListNode(-1) + } + + /** Get the value of the index-th node in the linked list. If the index is invalid, return -1. */ + func get(_ index: Int) -> Int { + if size > 0 && index < size { + var tempHead = head + for _ in 0.. size { + return + } + let idx = (index >= 0 ? index : 0) + var tempHead = head + for _ in 0 ..< idx { + tempHead = tempHead.next! + } + let currentNode = tempHead.next + let newNode = ListNode(val, currentNode) + tempHead.next = newNode + size += 1 + } + + /** Delete the index-th node in the linked list, if the index is valid. */ + func deleteAtIndex(_ index: Int) { + if size > 0 && index < size { + var tempHead = head + for _ in 0 ..< index { + tempHead = tempHead.next! + } + tempHead.next = tempHead.next!.next + size -= 1 + } + } +} +``` ----------------------- From 3f45f392f6c0235393171494790b43eea6350e30 Mon Sep 17 00:00:00 2001 From: X-shuffle <53906918+X-shuffle@users.noreply.github.com> Date: Tue, 17 Aug 2021 20:00:20 +0800 Subject: [PATCH 07/35] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200343.=E6=95=B4?= =?UTF-8?q?=E6=95=B0=E6=8B=86=E5=88=86=20go=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0343.整数拆分.md | 29 ++++++++++++++++++++++++++++- 1 file changed, 28 insertions(+), 1 deletion(-) diff --git a/problems/0343.整数拆分.md b/problems/0343.整数拆分.md index a9c55a85..03b6f0fc 100644 --- a/problems/0343.整数拆分.md +++ b/problems/0343.整数拆分.md @@ -227,7 +227,34 @@ class Solution: return dp[n] ``` Go： - +```golang +func integerBreak(n int) int { + /** + 动态五部曲 + 1.确定dp下标及其含义 + 2.确定递推公式 + 3.确定dp初始化 + 4.确定遍历顺序 + 5.打印dp + **/ + dp:=make([]int,n+1) + dp[1]=1 + dp[2]=1 + for i:=3;ib{ + return a + } + return b +} +``` Javascript: ```Javascript From cb9175f41a672e321cdd13ec5ba4a0fa7e3066da Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E4=BD=99=E6=9D=9C=E6=9E=97?= Date: Tue, 17 Aug 2021 22:53:03 +0800 Subject: [PATCH 08/35] =?UTF-8?q?=E6=B7=BB=E5=8A=A00707.=E8=AE=BE=E8=AE=A1?= =?UTF-8?q?=E9=93=BE=E8=A1=A8=20Swift=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0707.设计链表.md | 81 +++++++++++++++++++++++++++++++++++ 1 file changed, 81 insertions(+) diff --git a/problems/0707.设计链表.md b/problems/0707.设计链表.md index 0aa038e8..0b8cd2e4 100644 --- a/problems/0707.设计链表.md +++ b/problems/0707.设计链表.md @@ -948,6 +948,87 @@ class MyLinkedList { } ``` +Swift: + +```swift +class MyLinkedList { + var dummyHead: ListNode? + var size: Int + + init() { + dummyHead = ListNode(0) + size = 0 + } + + func get(_ index: Int) -> Int { + if index >= size || index < 0 { + return -1 + } + + var curNode = dummyHead?.next + var curIndex = index + + while curIndex > 0 { + curNode = curNode?.next + curIndex -= 1 + } + + return curNode?.value ?? -1 + } + + func addAtHead(_ val: Int) { + let newHead = ListNode(val) + newHead.next = dummyHead?.next + dummyHead?.next = newHead + size += 1 + } + + func addAtTail(_ val: Int) { + let newNode = ListNode(val) + var curNode = dummyHead + while curNode?.next != nil { + curNode = curNode?.next + } + + curNode?.next = newNode + size += 1 + } + + func addAtIndex(_ index: Int, _ val: Int) { + if index > size { + return + } + + let newNode = ListNode(val) + var curNode = dummyHead + var curIndex = index + + while curIndex > 0 { + curNode = curNode?.next + curIndex -= 1 + } + + newNode.next = curNode?.next + curNode?.next = newNode + size += 1 + } + + func deleteAtIndex(_ index: Int) { + if index >= size || index < 0 { + return + } + + var curNode = dummyHead + for _ in 0.. Date: Wed, 18 Aug 2021 01:26:55 +0800 Subject: [PATCH 09/35] =?UTF-8?q?Update=200102.=E4=BA=8C=E5=8F=89=E6=A0=91?= =?UTF-8?q?=E7=9A=84=E5=B1=82=E5=BA=8F=E9=81=8D=E5=8E=86.md?= 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 | 2 +- 1 file changed, 1 insertion(+), 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 0f9b2df6..9b1435a2 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 @@ -1129,7 +1129,7 @@ var largestValues = function(root) { queue.push(root); while(root!==null&&queue.length){ //设置max初始值就是队列的第一个元素 - let max=queue[0]; + let max=queue[0].val; let length=queue.length; while(length--){ let node = queue.shift(); From eaa0504acbdc195c0804ecaebd033528786e7939 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E6=9E=81=E5=AE=A2=E5=AD=A6=E4=BC=9F?= Date: Wed, 18 Aug 2021 19:47:30 +0800 Subject: [PATCH 10/35] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=20206.=E5=8F=8D?= =?UTF-8?q?=E8=BD=AC=E9=93=BE=E8=A1=A8=20Swift=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 | 37 +++++++++++++++++++++++++++++++++++ 1 file changed, 37 insertions(+) diff --git a/problems/0206.翻转链表.md b/problems/0206.翻转链表.md index 8bb359bd..0c850d52 100644 --- a/problems/0206.翻转链表.md +++ b/problems/0206.翻转链表.md @@ -334,6 +334,43 @@ fun reverseList(head: ListNode?): ListNode? { } ``` +Swift： +```swift +/// 双指针法 (迭代) +/// - Parameter head: 头结点 +/// - Returns: 翻转后的链表头结点 +func reverseList(_ head: ListNode?) -> ListNode? { + if head == nil || head?.next == nil { + return head + } + var pre: ListNode? = nil + var cur: ListNode? = head + var temp: ListNode? = nil + while cur != nil { + temp = cur?.next + cur?.next = pre + pre = cur + cur = temp + } + return pre +} + +/// 递归 +/// - Parameter head: 头结点 +/// - Returns: 翻转后的链表头结点 +func reverseList2(_ head: ListNode?) -> ListNode? { + return reverse(pre: nil, cur: head) +} +func reverse(pre: ListNode?, cur: ListNode?) -> ListNode? { + if cur == nil { + return pre + } + let temp: ListNode? = cur?.next + cur?.next = pre + return reverse(pre: cur, cur: temp) +} +``` + ----------------------- * 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 704c3bb7960fc40b3000d36e7ec6266747a9cc87 Mon Sep 17 00:00:00 2001 From: ironartisan Date: Wed, 18 Aug 2021 22:33:42 +0800 Subject: [PATCH 11/35] =?UTF-8?q?=E6=B7=BB=E5=8A=A00084.=E6=9F=B1=E7=8A=B6?= =?UTF-8?q?=E5=9B=BE=E4=B8=AD=E6=9C=80=E5=A4=A7=E7=9A=84=E7=9F=A9=E5=BD=A2?= =?UTF-8?q?java=E5=8A=A8=E6=80=81=E8=A7=84=E5=88=92=E8=A7=A3=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0084.柱状图中最大的矩形.md | 34 ++++++++++++++++++++ 1 file changed, 34 insertions(+) diff --git a/problems/0084.柱状图中最大的矩形.md b/problems/0084.柱状图中最大的矩形.md index 941888db..bec3bc99 100644 --- a/problems/0084.柱状图中最大的矩形.md +++ b/problems/0084.柱状图中最大的矩形.md @@ -195,6 +195,40 @@ public: Java: +动态规划 +```java +class Solution { + public int largestRectangleArea(int[] heights) { + int length = heights.length; + int[] minLeftIndex = new int [length]; + int[] maxRigthIndex = new int [length]; + + // 记录左边第一个小于该柱子的下标 + minLeftIndex[0] = -1 ; + for (int i = 1; i < length; i++) { + int t = i - 1; + // 这里不是用if，而是不断向右寻找的过程 + while (t >= 0 && heights[t] >= heights[i]) t = minLeftIndex[t]; + minLeftIndex[i] = t; + } + // 记录每个柱子 右边第一个小于该柱子的下标 + maxRigthIndex[length - 1] = length; + for (int i = length - 2; i >= 0; i--) { + int t = i + 1; + while(t < length && heights[t] >= heights[i]) t = maxRigthIndex[t]; + maxRigthIndex[i] = t; + } + // 求和 + int result = 0; + for (int i = 0; i < length; i++) { + int sum = heights[i] * (maxRigthIndex[i] - minLeftIndex[i] - 1); + result = Math.max(sum, result); + } + return result; + } +} +``` + Python: 动态规划 From 4b71311621baad2dc356dbe100e608435cc47bd8 Mon Sep 17 00:00:00 2001 From: shuwen Date: Thu, 19 Aug 2021 12:41:16 +0800 Subject: [PATCH 12/35] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=20=E5=B1=82=E5=BA=8F?= =?UTF-8?q?=E9=81=8D=E5=8E=86=E4=B8=AD=E7=9A=84=20104.=20=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?=20Python=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 | 23 +++++++++++++++++++++++ 1 file changed, 23 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 0f9b2df6..dc386684 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 @@ -1532,6 +1532,29 @@ Java： Python： +```python 3 +class Solution: + def maxDepth(self, root: TreeNode) -> int: + if root == None: + return 0 + + queue_ = [root] + result = [] + while queue_: + length = len(queue_) + sub = [] + for i in range(length): + cur = queue_.pop(0) + sub.append(cur.val) + #子节点入队列 + if cur.left: queue_.append(cur.left) + if cur.right: queue_.append(cur.right) + result.append(sub) + + + return len(result) +``` + Go： From bd3d65c41530e2f8268986ce9f4b13d889d5925d Mon Sep 17 00:00:00 2001 From: DoubleYellowIce <65336599+DoubleYellowIce@users.noreply.github.com> Date: Thu, 19 Aug 2021 14:34:15 +0800 Subject: [PATCH 13/35] =?UTF-8?q?Update=200188.=E4=B9=B0=E5=8D=96=E8=82=A1?= =?UTF-8?q?=E7=A5=A8=E7=9A=84=E6=9C=80=E4=BD=B3=E6=97=B6=E6=9C=BAIV.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../0188.买卖股票的最佳时机IV.md | 21 ++++++++++++++++++- 1 file changed, 20 insertions(+), 1 deletion(-) diff --git a/problems/0188.买卖股票的最佳时机IV.md b/problems/0188.买卖股票的最佳时机IV.md index 9fe7a919..66676a7a 100644 --- a/problems/0188.买卖股票的最佳时机IV.md +++ b/problems/0188.买卖股票的最佳时机IV.md @@ -196,7 +196,7 @@ class Solution { } } -// 版本二: 空间优化 +// 版本二: 二维 dp数组 class Solution { public int maxProfit(int k, int[] prices) { if (prices.length == 0) return 0; @@ -220,6 +220,25 @@ class Solution { return dp[len - 1][k*2]; } } + +//版本三：一维 dp数组 +class Solution { + public int maxProfit(int k, int[] prices) { + //在版本二的基础上，由于我们只关心前一天的股票买入情况，所以只存储前一天的股票买入情况 + if(prices.length==0)return 0; + int[] dp=new int[2*k+1]; + for (int i = 1; i <2*k ; i+=2) { + dp[i]=-prices[0]; + } + for (int i = 0; i Date: Thu, 19 Aug 2021 14:46:15 +0800 Subject: [PATCH 14/35] =?UTF-8?q?=E6=B7=BB=E5=8A=A00279.=E5=AE=8C=E5=85=A8?= =?UTF-8?q?=E5=B9=B3=E6=96=B9=E6=95=B0python3=E7=89=88=E6=9C=AC=E4=B8=80?= =?UTF-8?q?=E8=A7=A3=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0279.完全平方数.md | 16 ++++++++++++++++ 1 file changed, 16 insertions(+) diff --git a/problems/0279.完全平方数.md b/problems/0279.完全平方数.md index e51f0a99..863bd60c 100644 --- a/problems/0279.完全平方数.md +++ b/problems/0279.完全平方数.md @@ -217,6 +217,22 @@ class Solution: Python3: ```python class Solution: + '''版本一，先遍历背包, 再遍历物品''' + def numSquares(self, n: int) -> int: + dp = [n] * (n + 1) + dp[0] = 0 + # 遍历背包 + for j in range(1, n+1): + for i in range(1, n): + num = i ** 2 + if num > j: break + # 遍历物品 + if j - num >= 0: + dp[j] = min(dp[j], dp[j - num] + 1) + return dp[n] + +class Solution: + '''版本二， 先遍历物品, 再遍历背包''' def numSquares(self, n: int) -> int: # 初始化 # 组成和的完全平方数的最多个数，就是只用1构成 From 7c6930971d6175fe6656a4b31c4ac1c6d765d795 Mon Sep 17 00:00:00 2001 From: Eyjan_Huang <81480748+Eyjan-Huang@users.noreply.github.com> Date: Thu, 19 Aug 2021 15:00:28 +0800 Subject: [PATCH 15/35] =?UTF-8?q?=E6=9B=B4=E6=96=B0=200203.=E7=A7=BB?= =?UTF-8?q?=E9=99=A4=E9=93=BE=E8=A1=A8=E5=85=83=E7=B4=A0.md=20python?= =?UTF-8?q?=E4=BB=A3=E7=A0=81?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 使用更符合PEP8要求的代码规范来约束 --- problems/0203.移除链表元素.md | 10 ++++++---- 1 file changed, 6 insertions(+), 4 deletions(-) diff --git a/problems/0203.移除链表元素.md b/problems/0203.移除链表元素.md index f3724fc2..d67a7d2a 100644 --- a/problems/0203.移除链表元素.md +++ b/problems/0203.移除链表元素.md @@ -245,13 +245,15 @@ Python： # 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 + dummy_head = ListNode(next=head) cur = dummy_head - while(cur.next!=None): - if(cur.next.val == val): - cur.next = cur.next.next #删除cur.next节点 + + while cur.next: + if cur.next.val == val: + cur.next = cur.next.next # 删除下一个节点 else: cur = cur.next return dummy_head.next From aa4ce0f096013a20847b4b7c325a22fbe26e6deb Mon Sep 17 00:00:00 2001 From: ironartisan Date: Thu, 19 Aug 2021 15:03:44 +0800 Subject: [PATCH 16/35] =?UTF-8?q?=E6=B7=BB=E5=8A=A00279.=E5=AE=8C=E5=85=A8?= =?UTF-8?q?=E5=B9=B3=E6=96=B9=E6=95=B0java=E7=89=88=E6=9C=AC=E4=BA=8C?= =?UTF-8?q?=E8=A7=A3=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0279.完全平方数.md | 27 ++++++++++++++++++++++++++- 1 file changed, 26 insertions(+), 1 deletion(-) diff --git a/problems/0279.完全平方数.md b/problems/0279.完全平方数.md index 863bd60c..f23453da 100644 --- a/problems/0279.完全平方数.md +++ b/problems/0279.完全平方数.md @@ -161,6 +161,7 @@ public: Java： ```Java class Solution { + // 版本一，先遍历物品, 再遍历背包 public int numSquares(int n) { int max = Integer.MAX_VALUE; int[] dp = new int[n + 1]; @@ -170,7 +171,9 @@ class Solution { } //当和为0时，组合的个数为0 dp[0] = 0; + // 遍历物品 for (int i = 1; i * i <= n; i++) { + // 遍历背包 for (int j = i * i; j <= n; j++) { if (dp[j - i * i] != max) { dp[j] = Math.min(dp[j], dp[j - i * i] + 1); @@ -180,6 +183,28 @@ class Solution { return dp[n]; } } + +class Solution { + // 版本二， 先遍历背包, 再遍历物品 + public int numSquares(int n) { + int max = Integer.MAX_VALUE; + int[] dp = new int[n + 1]; + // 初始化 + for (int j = 0; j <= n; j++) { + dp[j] = max; + } + // 当和为0时，组合的个数为0 + dp[0] = 0; + // 遍历背包 + for (int j = 1; j <= n; j++) { + // 遍历物品 + for (int i = 1; i * i <= j; i++) { + dp[j] = Math.min(dp[j], dp[j - i * i] + 1); + } + } + return dp[n]; + } +} ``` Python： @@ -187,7 +212,7 @@ Python： ```python3 class Solution: def numSquares(self, n: int) -> int: - '''版本一''' + '''版本一，先遍历背包, 再遍历物品''' # 初始化 nums = [i**2 for i in range(1, n + 1) if i**2 <= n] dp = [10**4]*(n + 1) From 4df16b4258c41d3b6cfb0d9f7e4b9a51f9cba0a0 Mon Sep 17 00:00:00 2001 From: ironartisan Date: Thu, 19 Aug 2021 15:05:17 +0800 Subject: [PATCH 17/35] =?UTF-8?q?=E6=B7=BB=E5=8A=A00279.=E5=AE=8C=E5=85=A8?= =?UTF-8?q?=E5=B9=B3=E6=96=B9=E6=95=B0java=E7=89=88=E6=9C=AC=E4=BA=8C?= =?UTF-8?q?=E8=A7=A3=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0279.完全平方数.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/0279.完全平方数.md b/problems/0279.完全平方数.md index f23453da..3c0f0414 100644 --- a/problems/0279.完全平方数.md +++ b/problems/0279.完全平方数.md @@ -226,7 +226,7 @@ class Solution: return dp[n] def numSquares1(self, n: int) -> int: - '''版本二''' + '''版本二， 先遍历物品, 再遍历背包''' # 初始化 nums = [i**2 for i in range(1, n + 1) if i**2 <= n] dp = [10**4]*(n + 1) From f6eee532e49acba882be2b202825ee7082d9a69c Mon Sep 17 00:00:00 2001 From: Eyjan_Huang <81480748+Eyjan-Huang@users.noreply.github.com> Date: Thu, 19 Aug 2021 16:26:42 +0800 Subject: [PATCH 18/35] =?UTF-8?q?=E6=9B=B4=E6=96=B0=200344.=E5=8F=8D?= =?UTF-8?q?=E8=BD=AC=E5=AD=97=E7=AC=A6=E4=B8=B2.md=20python=E4=BB=A3?= =?UTF-8?q?=E7=A0=81=E4=BC=98=E5=8C=96=EF=BC=8C=E5=88=A0=E9=99=A4=E5=8F=A6?= =?UTF-8?q?=E4=B8=80=E7=A7=8D=E8=A7=A3=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 另一种解法进行了不必要的操作，且代码可读性较差。 推荐使用第一种解法 --- problems/0344.反转字符串.md | 17 +++++------------ 1 file changed, 5 insertions(+), 12 deletions(-) diff --git a/problems/0344.反转字符串.md b/problems/0344.反转字符串.md index 4f96f839..763097be 100644 --- a/problems/0344.反转字符串.md +++ b/problems/0344.反转字符串.md @@ -162,21 +162,14 @@ class Solution: Do not return anything, modify s in-place instead. """ left, right = 0, len(s) - 1 - while(left < right): + + # 该方法已经不需要判断奇偶数，经测试后时间空间复杂度比用 for i in range(right//2)更低 + # 推荐该写法，更加通俗易懂 + while left < right: s[left], s[right] = s[right], s[left] left += 1 right -= 1 - -# 下面的写法更加简洁，但是都是同样的算法 -# class Solution: -# def reverseString(self, s: List[str]) -> None: -# """ -# Do not return anything, modify s in-place instead. -# """ - # 不需要判别是偶数个还是奇数个序列，因为奇数个的时候，中间那个不需要交换就可 -# for i in range(len(s)//2): -# s[i], s[len(s)-1-i] = s[len(s)-1-i], s[i] -# return s + ``` Go： From 0990dfe810af83c4ed41c605fd165dad1f76645b Mon Sep 17 00:00:00 2001 From: shuwen Date: Thu, 19 Aug 2021 17:00:42 +0800 Subject: [PATCH 19/35] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200102.=E4=BA=8C?= =?UTF-8?q?=E5=8F=89=E6=A0=91=E7=9A=84=E5=B1=82=E5=BA=8F=E9=81=8D=E5=8E=86?= =?UTF-8?q?.md=20=20=E5=85=B3=E4=BA=8E111.=E4=BA=8C=E5=8F=89=E6=A0=91?= =?UTF-8?q?=E7=9A=84=E6=9C=80=E5=B0=8F=E6=B7=B1=E5=BA=A6=E7=9A=84Python?= =?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/0102.二叉树的层序遍历.md | 32 ++++++++++++++++++++++- 1 file changed, 31 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 dc386684..109ed70f 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 @@ -1562,6 +1562,8 @@ JavaScript： # 111.二叉树的最小深度 +题目地址：https://leetcode-cn.com/problems/minimum-depth-of-binary-tree/ + 相对于 104.二叉树的最大深度 ，本题还也可以使用层序遍历的方式来解决，思路是一样的。 **需要注意的是，只有当左右孩子都为空的时候，才说明遍历的最低点了。如果其中一个孩子为空则不是最低点** @@ -1597,7 +1599,35 @@ public: Java： -Python： +Python 3： + +```python 3 +# 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 minDepth(self, root: TreeNode) -> int: + if root == None: + return 0 + + #根节点的深度为1 + queue_ = [(root,1)] + while queue_: + cur, depth = queue_.pop(0) + + if cur.left == None and cur.right == None: + return depth + #先左子节点，由于左子节点没有孩子，则就是这一层了 + if cur.left: + queue_.append((cur.left,depth + 1)) + if cur.right: + queue_.append((cur.right,depth + 1)) + + return 0 +``` Go： From e6571ecaac55a91a557b32248560ef5fac612cf1 Mon Sep 17 00:00:00 2001 From: Eyjan_Huang <81480748+Eyjan-Huang@users.noreply.github.com> Date: Thu, 19 Aug 2021 17:30:36 +0800 Subject: [PATCH 20/35] =?UTF-8?q?=E6=9B=B4=E6=96=B0=200541.=E5=8F=8D?= =?UTF-8?q?=E8=BD=AC=E5=AD=97=E7=AC=A6=E4=B8=B2II.md=20python=E9=83=A8?= =?UTF-8?q?=E5=88=86=EF=BC=8C=E5=8E=BB=E9=99=A4=E8=87=AA=E9=80=A0=E8=BD=AE?= =?UTF-8?q?=E5=AD=90=E9=83=A8=E5=88=86=EF=BC=8C=E4=BC=98=E5=8C=96=E4=BB=A3?= =?UTF-8?q?=E7=A0=81=E5=8F=AF=E8=AF=BB=E6=80=A7?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 维持原方法的思路，简化了包和匿名函数的用法 原代码使用了reduce包 + 匿名函数来实现 ''.join的操作，去除该部分提升运行效率 --- problems/0541.反转字符串II.md | 35 ++++++++++++------------------ 1 file changed, 14 insertions(+), 21 deletions(-) diff --git a/problems/0541.反转字符串II.md b/problems/0541.反转字符串II.md index ab1ef16a..02713c65 100644 --- a/problems/0541.反转字符串II.md +++ b/problems/0541.反转字符串II.md @@ -155,34 +155,27 @@ class Solution { Python： ```python - -class Solution(object): - def reverseStr(self, s, k): +class Solution: + def reverseStr(self, s: str, k: int) -> str: """ - :type s: str - :type k: int - :rtype: str + 1. 使用range(start, end, step)来确定需要调换的初始位置 + 2. 对于字符串s = 'abc'，如果使用s[0:999] ===> 'abc'。字符串末尾如果超过最大长度，则会返回至字符串最后一个值，这个特性可以避免一些边界条件的处理。 + 3. 用切片整体替换，而不是一个个替换. """ - from functools import reduce - # turn s into a list - s = list(s) - - # another way to simply use a[::-1], but i feel this is easier to understand - def reverse(s): - left, right = 0, len(s) - 1 + def reverse_substring(text): + left, right = 0, len(text) - 1 while left < right: - s[left], s[right] = s[right], s[left] + text[left], text[right] = text[right], text[left] left += 1 right -= 1 - return s + return text - # make sure we reverse each 2k elements - for i in range(0, len(s), 2*k): - s[i:(i+k)] = reverse(s[i:(i+k)]) - - # combine list into str. - return reduce(lambda a, b: a+b, s) + res = list(s) + + for cur in range(0, len(s), 2 * k): + res[cur: cur + k] = reverse_substring(res[cur: cur + k]) + return ''.join(res) ``` From 51e719a4351fb97ed41ad6995bc6470d5204ad35 Mon Sep 17 00:00:00 2001 From: Eyjan_Huang <81480748+Eyjan-Huang@users.noreply.github.com> Date: Thu, 19 Aug 2021 17:57:39 +0800 Subject: [PATCH 21/35] =?UTF-8?q?=E7=AE=80=E5=8C=96=20=E5=89=91=E6=8C=87Of?= =?UTF-8?q?fer05.=E6=9B=BF=E6=8D=A2=E7=A9=BA=E6=A0=BC.md=20python=E4=BB=A3?= =?UTF-8?q?=E7=A0=81?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 减少不必要的变量使用，优化空间复杂度，使用切片替换而不是单个替换 --- problems/剑指Offer05.替换空格.md | 58 +++++++++----------------- 1 file changed, 20 insertions(+), 38 deletions(-) diff --git a/problems/剑指Offer05.替换空格.md b/problems/剑指Offer05.替换空格.md index 4ae5f9f2..3f373d0d 100644 --- a/problems/剑指Offer05.替换空格.md +++ b/problems/剑指Offer05.替换空格.md @@ -202,45 +202,27 @@ func replaceSpace(s string) string { python： ```python -class Solution(object): - def replaceSpace(self, s): - """ - :type s: str - :rtype: str - """ - list_s = list(s) - - # 记录原本字符串的长度 - original_end = len(s) - - # 将空格改成%20 使得字符串总长增长 2n，n为原本空格数量。 - # 所以记录空格数量就可以得到目标字符串的长度 - n_space = 0 - for ss in s: - if ss == ' ': - n_space += 1 - - list_s += ['0'] * 2 * n_space - - # 设置左右指针位置 - left, right = original_end - 1, len(list_s) - 1 - - # 循环直至左指针越界 - while left >= 0: - if list_s[left] == ' ': - list_s[right] = '0' - list_s[right - 1] = '2' - list_s[right - 2] = '%' - right -= 3 - else: - list_s[right] = list_s[left] - right -= 1 - - left -= 1 +class Solution: + def replaceSpace(self, s: str) -> str: + counter = s.count(' ') - # 将list变回str，输出 - s = ''.join(list_s) - return s + res = list(s) + # 每碰到一个空格就多拓展两个格子，1 + 2 = 3个位置存’%20‘ + res.extend([' '] * counter * 2) + + # 原始字符串的末尾，拓展后的末尾 + left, right = len(s) - 1, len(res) - 1 + + while left >= 0: + if res[left] != ' ': + res[right] = res[left] + right -= 1 + else: + # [right - 2, right), 左闭右开 + res[right - 2: right + 1] = '%20' + right -= 3 + left -= 1 + return ''.join(res) ``` From 6255eae04cefc676d243bc53ea4070bc900b3c99 Mon Sep 17 00:00:00 2001 From: Eyjan_Huang <81480748+Eyjan-Huang@users.noreply.github.com> Date: Thu, 19 Aug 2021 21:01:20 +0800 Subject: [PATCH 22/35] =?UTF-8?q?=E8=A1=A5=E5=85=85=20=E5=89=91=E6=8C=87Of?= =?UTF-8?q?fer58-II.=E5=B7=A6=E6=97=8B=E8=BD=AC=E5=AD=97=E7=AC=A6=E4=B8=B2?= =?UTF-8?q?.md=20python=E6=96=B9=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 如果不让使用自带函数reversed() 可以使用该方法 --- problems/剑指Offer58-II.左旋转字符串.md | 17 ++++++++++++++++- 1 file changed, 16 insertions(+), 1 deletion(-) 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 1073dafa..6070f85b 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 @@ -136,13 +136,28 @@ class Solution: # return "".join(s) +# 方法三：如果连reversed也不让使用，那么自己手写一个 +class Solution: + def reverseLeftWords(self, s: str, n: int) -> str: + def reverse_sub(lst, left, right): + while left < right: + lst[left], lst[right] = lst[right], lst[left] + left += 1 + right -= 1 + + res = list(s) + end = len(res) - 1 + reverse_sub(res, 0, n - 1) + reverse_sub(res, n, end) + reverse_sub(res, 0, end) + return ''.join(res) # 时间复杂度：O(n) # 空间复杂度：O(n)，python的string为不可变，需要开辟同样大小的list空间来修改 ``` ```python 3 -#方法三：考虑不能用切片的情况下，利用模+下标实现 +#方法四：考虑不能用切片的情况下，利用模+下标实现 class Solution: def reverseLeftWords(self, s: str, n: int) -> str: new_s = '' From 1a82f98a175ab03569c67a436b093203ab343d20 Mon Sep 17 00:00:00 2001 From: Eyjan_Huang <81480748+Eyjan-Huang@users.noreply.github.com> Date: Thu, 19 Aug 2021 21:04:57 +0800 Subject: [PATCH 23/35] =?UTF-8?q?=E6=9B=B4=E6=96=B0=20=E5=89=91=E6=8C=87Of?= =?UTF-8?q?fer58-II.=E5=B7=A6=E6=97=8B=E8=BD=AC=E5=AD=97=E7=AC=A6=E4=B8=B2?= =?UTF-8?q?.md=20=E6=A0=BC=E5=BC=8F=E4=BF=AE=E6=94=B9?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 更新排版 --- .../剑指Offer58-II.左旋转字符串.md | 22 ++++++++++++------- 1 file changed, 14 insertions(+), 8 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 6070f85b..15607a50 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 @@ -125,17 +125,21 @@ python: class Solution: def reverseLeftWords(self, s: str, n: int) -> str: return s[n:] + s[0:n] - +``` +```python # 方法二：也可以使用上文描述的方法，有些面试中不允许使用切片，那就使用上文作者提到的方法 -# class Solution: -# def reverseLeftWords(self, s: str, n: int) -> str: -# s = list(s) -# s[0:n] = list(reversed(s[0:n])) -# s[n:] = list(reversed(s[n:])) -# s.reverse() +class Solution: + def reverseLeftWords(self, s: str, n: int) -> str: + s = list(s) + s[0:n] = list(reversed(s[0:n])) + s[n:] = list(reversed(s[n:])) + s.reverse() + + return "".join(s) -# return "".join(s) +``` +```python # 方法三：如果连reversed也不让使用，那么自己手写一个 class Solution: def reverseLeftWords(self, s: str, n: int) -> str: @@ -152,8 +156,10 @@ class Solution: reverse_sub(res, 0, end) return ''.join(res) +# 同方法二 # 时间复杂度：O(n) # 空间复杂度：O(n)，python的string为不可变，需要开辟同样大小的list空间来修改 + ``` ```python 3 From bf92f955a3442d29c8c9f3e409d814ab644a63d1 Mon Sep 17 00:00:00 2001 From: X-shuffle <53906918+X-shuffle@users.noreply.github.com> Date: Thu, 19 Aug 2021 21:45:44 +0800 Subject: [PATCH 24/35] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=201002.=E6=9F=A5?= =?UTF-8?q?=E6=89=BE=E5=B8=B8=E7=94=A8=E5=AD=97=E7=AC=A6=20GO=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 1002.查找常用字符 GO版本 --- problems/1002.查找常用字符.md | 36 ++++++++++++++++++++++++++++- 1 file changed, 35 insertions(+), 1 deletion(-) diff --git a/problems/1002.查找常用字符.md b/problems/1002.查找常用字符.md index 723e5656..a789373b 100644 --- a/problems/1002.查找常用字符.md +++ b/problems/1002.查找常用字符.md @@ -233,7 +233,41 @@ var commonChars = function (words) { return res }; ``` - +GO +```golang +func commonChars(words []string) []string { + length:=len(words) + fre:=make([][]int,0)//统计每个字符串的词频 + res:=make([]string,0) + //统计词频 + for i:=0;ib{ + return b + } + return a +} +``` ----------------------- * 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 d51d819060914d89b4eb89bd9a9495e48a10ab22 Mon Sep 17 00:00:00 2001 From: X-shuffle <53906918+X-shuffle@users.noreply.github.com> Date: Thu, 19 Aug 2021 22:02:21 +0800 Subject: [PATCH 25/35] =?UTF-8?q?=E6=9B=B4=E6=96=B0=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=E6=9C=AC=EF=BC=88=E5=88=A9=E7=94=A8set=EF=BC=8C?= =?UTF-8?q?=E4=B8=8D=E7=94=A8=E7=BB=9F=E8=AE=A1=E6=AC=A1=E6=95=B0=EF=BC=8C?= =?UTF-8?q?=E5=87=8F=E5=B0=91=E7=A9=BA=E9=97=B4=E5=A4=8D=E6=9D=82=E5=BA=A6?= =?UTF-8?q?=EF=BC=89?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 更新 0349.两个数组的交集 go版本（利用set，不用统计次数，减少空间复杂度） --- problems/0349.两个数组的交集.md | 20 ++++++++++++++++++++ 1 file changed, 20 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 7489352d..62abf639 100644 --- a/problems/0349.两个数组的交集.md +++ b/problems/0349.两个数组的交集.md @@ -143,6 +143,26 @@ func intersection(nums1 []int, nums2 []int) []int { return res } ``` +```golang +//优化版，利用set，减少count统计 +func intersection(nums1 []int, nums2 []int) []int { + set:=make(map[int]struct{},0) + res:=make([]int,0) + for _,v:=range nums1{ + if _,ok:=set[v];!ok{ + set[v]=struct{}{} + } + } + for _,v:=range nums2{ + //如果存在于上一个数组中，则加入结果集，并清空该set值 + if _,ok:=set[v];ok{ + res=append(res,v) + delete(set, v) + } + } + return res +} +``` javaScript: From a69006f901230b1ba73e73ce8b86918b404480e4 Mon Sep 17 00:00:00 2001 From: Eyjan_Huang <81480748+Eyjan-Huang@users.noreply.github.com> Date: Fri, 20 Aug 2021 00:06:33 +0800 Subject: [PATCH 26/35] =?UTF-8?q?=E6=9B=B4=E6=96=B0=200232.=E7=94=A8?= =?UTF-8?q?=E6=A0=88=E5=AE=9E=E7=8E=B0=E9=98=9F=E5=88=97.md=20python?= =?UTF-8?q?=E4=BB=A3=E7=A0=81?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit python代码简化，符合PEP8标准，可读性加强，效率提升 --- problems/0232.用栈实现队列.md | 50 ++++++++++++++++------------- 1 file changed, 28 insertions(+), 22 deletions(-) diff --git a/problems/0232.用栈实现队列.md b/problems/0232.用栈实现队列.md index 0df82d35..46d884d3 100644 --- a/problems/0232.用栈实现队列.md +++ b/problems/0232.用栈实现队列.md @@ -281,48 +281,54 @@ 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. + in主要负责push，out主要负责pop """ - self.stack1 = list() - self.stack2 = list() + self.stack_in = [] + self.stack_out = [] + def push(self, x: int) -> None: """ - Push element x to the back of queue. + 有新元素进来，就往in里面push """ - # self.stack1用于接受元素 - self.stack1.append(x) + self.stack_in.append(x) + def pop(self) -> int: """ - Removes the element from in front of queue and returns that element. + 1. 检查如果out里面元素，则直接pop + 2. 如果out没有元素，就把in里面的元素（除了第一个）依次pop后装进out里面 + 3. 直接把in剩下的元素pop出来，就是queue头部的 """ - # 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() + if self.stack_out: + return self.stack_out.pop() + else: + for i in range(1, len(self.stack_in)): + self.stack_out.append(self.stack_in.pop()) + return self.stack_in.pop() + def peek(self) -> int: """ - Get the front element. + 1. 查out有没有元素，有就把最上面的返回 + 2. 如果out没有元素，就把in最下面的返回 """ - if self.stack2 == []: - while self.stack1: - tmp = self.stack1.pop() - self.stack2.append(tmp) - return self.stack2[-1] + if self.stack_out: + return self.stack_out[-1] + else: + return self.stack_in[0] + def empty(self) -> bool: """ - Returns whether the queue is empty. + 只要in或者out有元素，说明队列不为空 """ - return self.stack1 == [] and self.stack2 == [] + return not (self.stack_in or self.stack_out) + ``` From 77e789a248dc21b8b939ee37e0d5d58512e90d8d Mon Sep 17 00:00:00 2001 From: Eyjan_Huang <81480748+Eyjan-Huang@users.noreply.github.com> Date: Fri, 20 Aug 2021 01:49:06 +0800 Subject: [PATCH 27/35] =?UTF-8?q?=E7=AE=80=E5=8C=96=200225.=E7=94=A8?= =?UTF-8?q?=E9=98=9F=E5=88=97=E5=AE=9E=E7=8E=B0=E6=A0=88.md=20python?= =?UTF-8?q?=E9=83=A8=E5=88=86=EF=BC=8C=E5=86=97=E4=BD=99=E9=83=A8=E5=88=86?= =?UTF-8?q?=E5=A4=AA=E5=A4=9A?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 原方法用了很多不必要的变量和操作，改进后的代码更直观，效率和空间都得到了优化 --- problems/0225.用队列实现栈.md | 65 +++++++++++++++++------------ 1 file changed, 39 insertions(+), 26 deletions(-) diff --git a/problems/0225.用队列实现栈.md b/problems/0225.用队列实现栈.md index b327c17b..d82220db 100644 --- a/problems/0225.用队列实现栈.md +++ b/problems/0225.用队列实现栈.md @@ -294,53 +294,66 @@ Python： ```python from collections import deque + class MyStack: + def __init__(self): """ - Initialize your data structure here. + Python普通的Queue或SimpleQueue没有类似于peek的功能 + 也无法用索引访问，在实现top的时候较为困难。 + + 用list可以，但是在使用pop(0)的时候时间复杂度为O(1) + 因此这里使用双向队列，我们保证只执行popleft()和append()，因为deque可以用索引访问，可以实现和peek相似的功能 + + in - 存所有数据 + out - 仅在pop的时候会用到 """ - #使用两个队列来实现 - self.que1 = deque() - self.que2 = deque() + self.queue_in = deque() + self.queue_out = deque() def push(self, x: int) -> None: """ - Push element x onto stack. + 直接append即可 """ - self.que1.append(x) + self.queue_in.append(x) + def pop(self) -> int: """ - Removes the element on top of the stack and returns that element. + 1. 首先确认不空 + 2. 因为队列的特殊性，FIFO，所以我们只有在pop()的时候才会使用queue_out + 3. 先把queue_in中的所有元素（除了最后一个），依次出列放进queue_out + 4. 交换in和out，此时out里只有一个元素 + 5. 把out中的pop出来，即是原队列的最后一个 + + tip：这不能像栈实现队列一样，因为另一个queue也是FIFO，如果执行pop()它不能像 + stack一样从另一个pop()，所以干脆in只用来存数据，pop()的时候两个进行交换 """ - size = len(self.que1) - size -= 1#这里先减一是为了保证最后面的元素 - while size > 0: - size -= 1 - self.que2.append(self.que1.popleft()) + if self.empty(): + return None - - result = self.que1.popleft() - self.que1, self.que2= self.que2, self.que1#将que2和que1交换 que1经过之前的操作应该是空了 - #一定注意不能直接使用que1 = que2 这样que2的改变会影响que1 可以用浅拷贝 - return result + for i in range(len(self.queue_in) - 1): + self.queue_out.append(self.queue_in.popleft()) + + self.queue_in, self.queue_out = self.queue_out, self.queue_in # 交换in和out，这也是为啥in只用来存 + return self.queue_out.popleft() def top(self) -> int: """ - Get the top element. + 1. 首先确认不空 + 2. 我们仅有in会存放数据，所以返回第一个即可 """ - return self.que1[-1] + if self.empty(): + return None + + return self.queue_in[-1] + def empty(self) -> bool: """ - Returns whether the stack is empty. + 因为只有in存了数据，只要判断in是不是有数即可 """ - #print(self.que1) - if len(self.que1) == 0: - return True - else: - return False - + return len(self.queue_in) == 0 ``` From 55eb649434a95116e0fdcde2a89db5be1f85d801 Mon Sep 17 00:00:00 2001 From: Eyjan_Huang <81480748+Eyjan-Huang@users.noreply.github.com> Date: Fri, 20 Aug 2021 01:50:53 +0800 Subject: [PATCH 28/35] =?UTF-8?q?=E6=9B=B4=E6=AD=A3=200225.=E7=94=A8?= =?UTF-8?q?=E9=98=9F=E5=88=97=E5=AE=9E=E7=8E=B0=E6=A0=88.md=EF=BC=8Cdoc?= =?UTF-8?q?=E6=96=87=E6=A1=A3=E9=94=99=E8=AF=AF?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 时间复杂度错误，修正 --- problems/0225.用队列实现栈.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/0225.用队列实现栈.md b/problems/0225.用队列实现栈.md index d82220db..ccf93f1f 100644 --- a/problems/0225.用队列实现栈.md +++ b/problems/0225.用队列实现栈.md @@ -302,7 +302,7 @@ class MyStack: Python普通的Queue或SimpleQueue没有类似于peek的功能 也无法用索引访问，在实现top的时候较为困难。 - 用list可以，但是在使用pop(0)的时候时间复杂度为O(1) + 用list可以，但是在使用pop(0)的时候时间复杂度为O(n) 因此这里使用双向队列，我们保证只执行popleft()和append()，因为deque可以用索引访问，可以实现和peek相似的功能 in - 存所有数据 From 571defa00789aa9764abb6aa7efe54944ed42a49 Mon Sep 17 00:00:00 2001 From: Eyjan_Huang <81480748+Eyjan-Huang@users.noreply.github.com> Date: Fri, 20 Aug 2021 02:15:03 +0800 Subject: [PATCH 29/35] =?UTF-8?q?=E6=9B=B4=E6=96=B0=200020.=E6=9C=89?= =?UTF-8?q?=E6=95=88=E7=9A=84=E6=8B=AC=E5=8F=B7.md=20python=E9=83=A8?= =?UTF-8?q?=E5=88=86=E9=A2=9D=E5=A4=96=E6=96=B9=E6=B3=95=E6=8F=90=E4=BE=9B?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 补充了题主的方法，改写了原先冗余的做法，可读性提升，PEP8标准 --- problems/0020.有效的括号.md | 44 +++++++++++++++++++++++++------- 1 file changed, 35 insertions(+), 9 deletions(-) diff --git a/problems/0020.有效的括号.md b/problems/0020.有效的括号.md index a26aa308..57729c38 100644 --- a/problems/0020.有效的括号.md +++ b/problems/0020.有效的括号.md @@ -162,18 +162,44 @@ class Solution { Python： ```python3 +# 方法一，仅使用栈，更省空间 class Solution: def isValid(self, s: str) -> bool: - stack = [] # 保存还未匹配的左括号 - mapping = {")": "(", "]": "[", "}": "{"} - for i in s: - if i in "([{": # 当前是左括号，则入栈 - stack.append(i) - elif stack and stack[-1] == mapping[i]: # 当前是配对的右括号则出栈 - stack.pop() - else: # 不是匹配的右括号或者没有左括号与之匹配，则返回false + stack = [] + + for item in s: + if item == '(': + stack.append(')') + elif item == '[': + stack.append(']') + elif item == '{': + stack.append('}') + elif not stack or stack[-1] != item: return False - return stack == [] # 最后必须正好把左括号匹配完 + else: + stack.pop() + + return True if not stack else False +``` + +```python3 +# 方法二，使用字典 +class Solution: + def isValid(self, s: str) -> bool: + stack = [] + mapping = { + '(': ')', + '[': ']', + '{': '}' + } + for item in s: + if item in mapping.keys(): + stack.append(mapping[item]) + elif not stack or stack[-1] != item: + return False + else: + stack.pop() + return True if not stack else False ``` Go： From ce2c6fb562ae22b2b18230d8f00202f67c89c696 Mon Sep 17 00:00:00 2001 From: Eyjan_Huang <81480748+Eyjan-Huang@users.noreply.github.com> Date: Fri, 20 Aug 2021 02:55:31 +0800 Subject: [PATCH 30/35] =?UTF-8?q?=E6=9B=B4=E6=96=B0+=E8=A1=A5=E5=85=85=201?= =?UTF-8?q?047.=E5=88=A0=E9=99=A4=E5=AD=97=E7=AC=A6=E4=B8=B2=E4=B8=AD?= =?UTF-8?q?=E7=9A=84=E6=89=80=E6=9C=89=E7=9B=B8=E9=82=BB=E9=87=8D=E5=A4=8D?= =?UTF-8?q?=E9=A1=B9.md=20python=E4=BB=A3=E7=A0=81?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 补充使用双指针的方法，但是更推荐栈的做法 --- ...除字符串中的所有相邻重复项.md | 35 +++++++++++++++---- 1 file changed, 29 insertions(+), 6 deletions(-) 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 c4ae85c9..7b059d40 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 @@ -197,15 +197,38 @@ class Solution { Python： ```python3 +# 方法一，使用栈，推荐！ class Solution: def removeDuplicates(self, s: str) -> str: - t = list() - for i in s: - if t and t[-1] == i: - t.pop(-1) + res = list() + for item in s: + if res and res[-1] == item: + t.pop() else: - t.append(i) - return "".join(t) # 字符串拼接 + t.append(item) + return "".join(res) # 字符串拼接 +``` + +```python3 +# 双指针 +class Solution: + def removeDuplicates(self, s: str) -> str: + res = list(s) + slow = fast = 0 + length = len(res) + + while fast < length: + # 如果一样直接换，不一样会把后面的填在slow的位置 + res[slow] = res[fast] + + # 如果发现和前一个一样，就退一格指针 + if slow > 0 and res[slow] == res[slow - 1]: + slow -= 1 + else: + slow += 1 + fast += 1 + + return ''.join(res[0: slow]) ``` Go： From 13293e8c2f0e30743848e632244cc1a112e0422c Mon Sep 17 00:00:00 2001 From: Eyjan_Huang <81480748+Eyjan-Huang@users.noreply.github.com> Date: Fri, 20 Aug 2021 02:58:09 +0800 Subject: [PATCH 31/35] =?UTF-8?q?=E6=9B=B4=E6=96=B0=201047.=E5=88=A0?= =?UTF-8?q?=E9=99=A4=E5=AD=97=E7=AC=A6=E4=B8=B2=E4=B8=AD=E7=9A=84=E6=89=80?= =?UTF-8?q?=E6=9C=89=E7=9B=B8=E9=82=BB=E9=87=8D=E5=A4=8D=E9=A1=B9.md=20pyt?= =?UTF-8?q?hon=E6=B3=A8=E9=87=8A?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/1047.删除字符串中的所有相邻重复项.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) 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 7b059d40..1d63e7f8 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 @@ -210,7 +210,7 @@ class Solution: ``` ```python3 -# 双指针 +# 方法二，使用双指针模拟栈，如果不让用栈可以作为备选方法。 class Solution: def removeDuplicates(self, s: str) -> str: res = list(s) From 5e3d23fa38c78b9e66f12bd0e768968d7f31ed6c Mon Sep 17 00:00:00 2001 From: Eyjan_Huang <81480748+Eyjan-Huang@users.noreply.github.com> Date: Fri, 20 Aug 2021 02:59:10 +0800 Subject: [PATCH 32/35] =?UTF-8?q?=E6=9B=B4=E6=AD=A3=201047.=E5=88=A0?= =?UTF-8?q?=E9=99=A4=E5=AD=97=E7=AC=A6=E4=B8=B2=E4=B8=AD=E7=9A=84=E6=89=80?= =?UTF-8?q?=E6=9C=89=E7=9B=B8=E9=82=BB=E9=87=8D=E5=A4=8D=E9=A1=B9.md=20pyt?= =?UTF-8?q?hon=E6=96=B9=E6=B3=95=E4=B8=80=E4=BB=A3=E7=A0=81=E7=9A=84?= =?UTF-8?q?=E9=94=99=E8=AF=AF?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/1047.删除字符串中的所有相邻重复项.md | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) 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 1d63e7f8..b60a8d1d 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 @@ -203,9 +203,9 @@ class Solution: res = list() for item in s: if res and res[-1] == item: - t.pop() + res.pop() else: - t.append(item) + res.append(item) return "".join(res) # 字符串拼接 ``` From 8b94306cc6030eb79f2a8376b01f6398d888f7ed Mon Sep 17 00:00:00 2001 From: Eyjan_Huang <81480748+Eyjan-Huang@users.noreply.github.com> Date: Fri, 20 Aug 2021 03:23:30 +0800 Subject: [PATCH 33/35] =?UTF-8?q?=E6=9B=B4=E6=96=B0=200024.=E4=B8=A4?= =?UTF-8?q?=E4=B8=A4=E4=BA=A4=E6=8D=A2=E9=93=BE=E8=A1=A8=E4=B8=AD=E7=9A=84?= =?UTF-8?q?=E8=8A=82=E7=82=B9.md=20python=E4=BB=A3=E7=A0=81?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 使用变量名使交换的过程更加清晰，而不是使用之前的next.next这样可读性较差 从后往前换 pre = dummy，cur = 2， post=3 dummpy -> 1 -> 2 -> None 先把cur.next = post.next，此时链表为 1 -> None 再 post.next = cur 此时链表为 2 -> 1 -> None, dummpy -> 1 -> None 最后 pre.next = post, 此时为 dummpy -> 2 -> 1 -> None --- .../0024.两两交换链表中的节点.md | 32 ++++++++++++------- 1 file changed, 20 insertions(+), 12 deletions(-) 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 91e566dd..672b9a8f 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 @@ -160,21 +160,29 @@ 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 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 + res = ListNode(next=head) + pre = res + + # 必须有pre的下一个和下下个才能交换，否则说明已经交换结束了 + while pre.next and pre.next.next: + cur = pre.next + post = pre.next.next - 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 + # pre，cur，post对应最左，中间的，最右边的节点 + cur.next = post.next + post.next = cur + pre.next = post + + pre = pre.next.next + return res.next ``` Go： From ca576c5bd0a62769c4b080028e7092d7fb945890 Mon Sep 17 00:00:00 2001 From: Eyjan_Huang <81480748+Eyjan-Huang@users.noreply.github.com> Date: Fri, 20 Aug 2021 03:29:26 +0800 Subject: [PATCH 34/35] =?UTF-8?q?=E6=9B=B4=E6=AD=A3=200232.=E7=94=A8?= =?UTF-8?q?=E6=A0=88=E5=AE=9E=E7=8E=B0=E9=98=9F=E5=88=97.md=20python?= =?UTF-8?q?=E4=BB=A3=E7=A0=81?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 添加限制条件，仅在栈不为空的情况下才允许后续pop()和peek()的操作 --- problems/0232.用栈实现队列.md | 6 ++++++ 1 file changed, 6 insertions(+) diff --git a/problems/0232.用栈实现队列.md b/problems/0232.用栈实现队列.md index 46d884d3..27a3de4e 100644 --- a/problems/0232.用栈实现队列.md +++ b/problems/0232.用栈实现队列.md @@ -304,6 +304,9 @@ class MyQueue: 2. 如果out没有元素，就把in里面的元素（除了第一个）依次pop后装进out里面 3. 直接把in剩下的元素pop出来，就是queue头部的 """ + if self.empty: + return None + if self.stack_out: return self.stack_out.pop() else: @@ -317,6 +320,9 @@ class MyQueue: 1. 查out有没有元素，有就把最上面的返回 2. 如果out没有元素，就把in最下面的返回 """ + if self.empty: + return None + if self.stack_out: return self.stack_out[-1] else: From 02fcd2c4a8f12782529b0fa787d3fd3f1f0fdca7 Mon Sep 17 00:00:00 2001 From: Eyjan_Huang <81480748+Eyjan-Huang@users.noreply.github.com> Date: Fri, 20 Aug 2021 17:14:16 +0800 Subject: [PATCH 35/35] =?UTF-8?q?=E6=9B=B4=E6=96=B0=200150.=E9=80=86?= =?UTF-8?q?=E6=B3=A2=E5=85=B0=E8=A1=A8=E8=BE=BE=E5=BC=8F=E6=B1=82=E5=80=BC?= =?UTF-8?q?.md=20python=E4=BB=A3=E7=A0=81?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 用format string提升效率，增加可读性，避免使用索引访问，直接使用切片。 --- problems/0150.逆波兰表达式求值.md | 24 ++++++++++++----------- 1 file changed, 13 insertions(+), 11 deletions(-) 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 2b294337..bcde7d5b 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 @@ -223,17 +223,19 @@ 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] +class Solution: + def evalRPN(self, tokens: List[str]) -> int: + stack = [] + for item in tokens: + if item not in {"+", "-", "*", "/"}: + stack.append(item) + else: + first_num, second_num = stack.pop(), stack.pop() + stack.append( + int(eval(f'{second_num} {item} {first_num}')) # 第一个出来的在运算符后面 + ) + return int(stack.pop()) # 如果一开始只有一个数，那么会是字符串形式的 + ```