From 0cfe55180439c77995abefaa02c5b61108203f3a Mon Sep 17 00:00:00 2001 From: asxy <17375702582@163.com> Date: Thu, 20 Apr 2023 11:14:43 +0800 Subject: [PATCH 01/76] =?UTF-8?q?Update=200647.=E5=9B=9E=E6=96=87=E5=AD=90?= =?UTF-8?q?=E4=B8=B2.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 回文子串添加java动态规划简单版本的代码 --- problems/0647.回文子串.md | 21 +++++++++++++++++++++ 1 file changed, 21 insertions(+) diff --git a/problems/0647.回文子串.md b/problems/0647.回文子串.md index 90e6da9f..521c8c26 100644 --- a/problems/0647.回文子串.md +++ b/problems/0647.回文子串.md @@ -267,6 +267,27 @@ class Solution { return ans; } } + +``` + +动态规划：简洁版 +```java +class Solution { + public int countSubstrings(String s) { + boolean[][] dp = new boolean[s.length()][s.length()]; + + int res = 0; + for (int i = s.length() - 1; i >= 0; i--) { + for (int j = i; j < s.length(); j++) { + if (s.charAt(i) == s.charAt(j) && (j - i <= 1 || dp[i + 1][j - 1])) { + res++; + dp[i][j] = true; + } + } + } + return res; + } +} ``` 中心扩散法： From fca305039005936bc9678c01d7573d6a2f8552ba Mon Sep 17 00:00:00 2001 From: asxy Date: Thu, 20 Apr 2023 11:47:39 +0800 Subject: [PATCH 02/76] =?UTF-8?q?Update=200084.=E6=9F=B1=E7=8A=B6=E5=9B=BE?= =?UTF-8?q?=E4=B8=AD=E6=9C=80=E5=A4=A7=E7=9A=84=E7=9F=A9=E5=BD=A2.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 添加单调栈精简java代码 --- problems/0084.柱状图中最大的矩形.md | 27 ++++++++++++++++++++ 1 file changed, 27 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 eb064143..f9a83508 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 @@ -307,6 +307,33 @@ class Solution { } } ``` +单调栈精简 +```java +class Solution { + public int largestRectangleArea(int[] heights) { + int[] newHeight = new int[heights.length + 2]; + System.arraycopy(heights, 0, newHeight, 1, heights.length); + newHeight[heights.length+1] = 0; + newHeight[0] = 0; + + Stack stack = new Stack<>(); + stack.push(0); + + int res = 0; + for (int i = 1; i < newHeight.length; i++) { + while (newHeight[i] < newHeight[stack.peek()]) { + int mid = stack.pop(); + int w = i - stack.peek() - 1; + int h = newHeight[mid]; + res = Math.max(res, w * h); + } + stack.push(i); + + } + return res; + } +} +``` Python3: From f7bdc58b2802e8b81483a5c2f8023499a3d525c4 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Thu, 20 Apr 2023 01:18:03 -0500 Subject: [PATCH 03/76] =?UTF-8?q?Update=200707.=E8=AE=BE=E8=AE=A1=E9=93=BE?= =?UTF-8?q?=E8=A1=A8.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 修改 python 代码，dummy node --- problems/0707.设计链表.md | 71 +++++++++++++++++++---------------- 1 file changed, 38 insertions(+), 33 deletions(-) diff --git a/problems/0707.设计链表.md b/problems/0707.设计链表.md index 5c78a12a..c72b7327 100644 --- a/problems/0707.设计链表.md +++ b/problems/0707.设计链表.md @@ -486,15 +486,10 @@ class MyLinkedList { Python： ```python # 单链表 -class Node(object): - def __init__(self, x=0): - self.val = x - self.next = None - -class MyLinkedList(object): +class MyLinkedList1: def __init__(self): - self.head = Node() + self.dummy_head = Node()# 添加虚拟头指针，便于操作 self.size = 0 # 设置一个链表长度的属性，便于后续操作，注意每次增和删的时候都要更新 def get(self, index): @@ -504,7 +499,7 @@ class MyLinkedList(object): """ if index < 0 or index >= self.size: return -1 - cur = self.head.next + cur = self.dummy_head.next while(index): cur = cur.next index -= 1 @@ -516,8 +511,8 @@ class MyLinkedList(object): :rtype: None """ new_node = Node(val) - new_node.next = self.head.next - self.head.next = new_node + new_node.next = self.dummy_head.next + self.dummy_head.next = new_node self.size += 1 def addAtTail(self, val): @@ -526,7 +521,7 @@ class MyLinkedList(object): :rtype: None """ new_node = Node(val) - cur = self.head + cur = self.dummy_head while(cur.next): cur = cur.next cur.next = new_node @@ -548,12 +543,12 @@ class MyLinkedList(object): return node = Node(val) - pre = self.head + cur = self.dummy_head while(index): - pre = pre.next + cur = cur.next index -= 1 - node.next = pre.next - pre.next = node + node.next = cur.next + cur.next = node self.size += 1 def deleteAtIndex(self, index): @@ -563,7 +558,7 @@ class MyLinkedList(object): """ if index < 0 or index >= self.size: return - pre = self.head + pre = self.dummy_head while(index): pre = pre.next index -= 1 @@ -574,11 +569,10 @@ class MyLinkedList(object): # 相对于单链表, Node新增了prev属性 class Node: - def __init__(self, val): + def __init__(self, val=0, next = None, prev = None): self.val = val - self.prev = None - self.next = None - + self.next = next + self.prev = prev class MyLinkedList: @@ -601,6 +595,20 @@ class MyLinkedList: node = node.next return node + + def _update(self, prev: Node, next: Node, val: int) -> None: + """ + 更新节点 + :param prev: 相对于更新的前一个节点 + :param next: 相对于更新的后一个节点 + :param val: 要添加的节点值 + """ + # 计数累加 + self._count += 1 + node = Node(val) + prev.next, next.prev = node, node + node.prev, node.next = prev, next + def get(self, index: int) -> int: """ Get the value of the index-th node in the linked list. If the index is invalid, return -1. @@ -634,19 +642,6 @@ class MyLinkedList: node = self._get_node(index) self._update(node.prev, node, val) - def _update(self, prev: Node, next: Node, val: int) -> None: - """ - 更新节点 - :param prev: 相对于更新的前一个节点 - :param next: 相对于更新的后一个节点 - :param val: 要添加的节点值 - """ - # 计数累加 - self._count += 1 - node = Node(val) - prev.next, next.prev = node, node - node.prev, node.next = prev, next - def deleteAtIndex(self, index: int) -> None: """ Delete the index-th node in the linked list, if the index is valid. @@ -656,6 +651,16 @@ class MyLinkedList: # 计数-1 self._count -= 1 node.prev.next, node.next.prev = node.next, node.prev + + + +# Your MyLinkedList object will be instantiated and called as such: +# obj = MyLinkedList() +# param_1 = obj.get(index) +# obj.addAtHead(val) +# obj.addAtTail(val) +# obj.addAtIndex(index,val) +# obj.deleteAtIndex(index) ``` Go： From 252e330a6b20fa13e81e4412c91f40547e0ba1b1 Mon Sep 17 00:00:00 2001 From: HOUSHENGREN <48871516+HOUSHENGREN@users.noreply.github.com> Date: Sun, 23 Apr 2023 14:36:27 +0800 Subject: [PATCH 04/76] =?UTF-8?q?Update=20=E5=88=B7=E5=8A=9B=E6=89=A3?= =?UTF-8?q?=E7=94=A8=E4=B8=8D=E7=94=A8=E5=BA=93=E5=87=BD=E6=95=B0.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit fix：文字错误 --- problems/前序/刷力扣用不用库函数.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/前序/刷力扣用不用库函数.md b/problems/前序/刷力扣用不用库函数.md index ae0940bf..04fce856 100644 --- a/problems/前序/刷力扣用不用库函数.md +++ b/problems/前序/刷力扣用不用库函数.md @@ -24,5 +24,5 @@ 例如for循环里套一个字符串的insert，erase之类的操作，你说时间复杂度是多少呢，很明显是O(n^2)的时间复杂度了。 -在刷题的时候本着我说的标准来使用库函数，详细对大家回有所帮助！ +在刷题的时候本着我说的标准来使用库函数，相信对大家回有所帮助！ From b8f6df60c432fd85d9417e0ce138a90b4603527e Mon Sep 17 00:00:00 2001 From: Lozakaka <102352821+Lozakaka@users.noreply.github.com> Date: Sun, 23 Apr 2023 02:40:34 -0400 Subject: [PATCH 05/76] =?UTF-8?q?Update=200106.=E4=BB=8E=E4=B8=AD=E5=BA=8F?= =?UTF-8?q?=E4=B8=8E=E5=90=8E=E5=BA=8F=E9=81=8D=E5=8E=86=E5=BA=8F=E5=88=97?= =?UTF-8?q?=E6=9E=84=E9=80=A0=E4=BA=8C=E5=8F=89=E6=A0=91.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 新增一java寫法 和卡哥的思路一樣的 --- ...序与后序遍历序列构造二叉树.md | 35 +++++++++++++++++++ 1 file changed, 35 insertions(+) diff --git a/problems/0106.从中序与后序遍历序列构造二叉树.md b/problems/0106.从中序与后序遍历序列构造二叉树.md index adb374f9..8fc973e0 100644 --- a/problems/0106.从中序与后序遍历序列构造二叉树.md +++ b/problems/0106.从中序与后序遍历序列构造二叉树.md @@ -622,7 +622,42 @@ class Solution { } } ``` +```java +class Solution { + public TreeNode buildTree(int[] inorder, int[] postorder) { + if(postorder.length == 0 || inorder.length == 0) + return null; + return buildHelper(inorder, 0, inorder.length, postorder, 0, postorder.length); + + } + private TreeNode buildHelper(int[] inorder, int inorderStart, int inorderEnd, int[] postorder, int postorderStart, int postorderEnd){ + if(postorderStart == postorderEnd) + return null; + int rootVal = postorder[postorderEnd - 1]; + TreeNode root = new TreeNode(rootVal); + int middleIndex; + for (middleIndex = inorderStart; middleIndex < inorderEnd; middleIndex++){ + if(inorder[middleIndex] == rootVal) + break; + } + int leftInorderStart = inorderStart; + int leftInorderEnd = middleIndex; + int rightInorderStart = middleIndex + 1; + int rightInorderEnd = inorderEnd; + + + int leftPostorderStart = postorderStart; + int leftPostorderEnd = postorderStart + (middleIndex - inorderStart); + int rightPostorderStart = leftPostorderEnd; + int rightPostorderEnd = postorderEnd - 1; + root.left = buildHelper(inorder, leftInorderStart, leftInorderEnd, postorder, leftPostorderStart, leftPostorderEnd); + root.right = buildHelper(inorder, rightInorderStart, rightInorderEnd, postorder, rightPostorderStart, rightPostorderEnd); + + return root; + } +} +``` 105.从前序与中序遍历序列构造二叉树 ```java From f17c74ceec294a90f864a1a9a3836a03b4a752bb Mon Sep 17 00:00:00 2001 From: HOUSHENGREN <48871516+HOUSHENGREN@users.noreply.github.com> Date: Sun, 23 Apr 2023 15:14:25 +0800 Subject: [PATCH 06/76] =?UTF-8?q?Update=20=E5=88=B7=E5=8A=9B=E6=89=A3?= =?UTF-8?q?=E7=94=A8=E4=B8=8D=E7=94=A8=E5=BA=93=E5=87=BD=E6=95=B0.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit feat：修改文案 --- problems/前序/刷力扣用不用库函数.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/前序/刷力扣用不用库函数.md b/problems/前序/刷力扣用不用库函数.md index 04fce856..7d0e3475 100644 --- a/problems/前序/刷力扣用不用库函数.md +++ b/problems/前序/刷力扣用不用库函数.md @@ -24,5 +24,5 @@ 例如for循环里套一个字符串的insert，erase之类的操作，你说时间复杂度是多少呢，很明显是O(n^2)的时间复杂度了。 -在刷题的时候本着我说的标准来使用库函数，相信对大家回有所帮助！ +在刷题的时候本着我说的标准来使用库函数，相信对大家会有所帮助！ From 9617a8b3fedc3996645977df31104159864b3462 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Mon, 1 May 2023 15:46:15 -0500 Subject: [PATCH 07/76] =?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 | 75 +++++++++++++---------- 1 file changed, 44 insertions(+), 31 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 26f0ae2d..e777c2a4 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 @@ -171,47 +171,60 @@ python3代码： ```python +# 利用长度法 +# 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 +from collections import deque class Solution: - """二叉树层序遍历迭代解法""" - - def levelOrder(self, root: TreeNode) -> List[List[int]]: - results = [] + def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]: if not root: - return results - - from collections import deque - que = deque([root]) - - while que: - size = len(que) - result = [] - for _ in range(size): - cur = que.popleft() - result.append(cur.val) + return [] + queue = deque([root]) + result = [] + while queue: + level = [] + for _ in range(len(queue)): + cur = queue.popleft() + level.append(cur.val) if cur.left: - que.append(cur.left) + queue.append(cur.left) if cur.right: - que.append(cur.right) - results.append(result) - - return results + queue.append(cur.right) + result.append(level) + return result ``` - ```python # 递归法 +# 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]]: - res = [] - def helper(root, depth): - if not root: return [] - if len(res) == depth: res.append([]) # start the current depth - res[depth].append(root.val) # fulfil the current depth - if root.left: helper(root.left, depth + 1) # process child nodes for the next depth - if root.right: helper(root.right, depth + 1) - helper(root, 0) - return res + def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]: + levels = [] + self.helper(root, 0, levels) + return levels + + def helper(self, node, level, levels): + if not node: + return + if len(levels) == level: + levels.append([]) + levels[level].append(node.val) + self.helper(node.left, level + 1, levels) + self.helper(node.right, level + 1, levels) + + ``` + + go: ```go From 424d5c224f229aaee353b3bbae8996b8e5dcbef2 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 15:06:11 -0500 Subject: [PATCH 08/76] =?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 | 17 ++++++++++++++++- 1 file changed, 16 insertions(+), 1 deletion(-) diff --git a/problems/0027.移除元素.md b/problems/0027.移除元素.md index 91150c74..6dc6404e 100644 --- a/problems/0027.移除元素.md +++ b/problems/0027.移除元素.md @@ -198,6 +198,7 @@ Python： ``` python 3 +（版本一）快慢指针法 class Solution: def removeElement(self, nums: List[int], val: int) -> int: # 快慢指针 @@ -213,7 +214,21 @@ class Solution: return slow ``` - +``` python 3 +（版本二）暴力递归法 +class Solution: + def removeElement(self, nums: List[int], val: int) -> int: + i, l = 0, len(nums) + while i < l: + if nums[i] == val: # 找到等于目标值的节点 + for j in range(i+1, l): # 移除该元素，并将后面元素向前平移 + nums[j - 1] = nums[j] + l -= 1 + i -= 1 + i += 1 + return l + +``` Go： From 53fbfc8339a6b61cb708ec87356e764db0f2b1ba Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 15:08:17 -0500 Subject: [PATCH 09/76] =?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 | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/0027.移除元素.md b/problems/0027.移除元素.md index 6dc6404e..90801153 100644 --- a/problems/0027.移除元素.md +++ b/problems/0027.移除元素.md @@ -215,7 +215,7 @@ class Solution: ``` ``` python 3 -（版本二）暴力递归法 +（版本二）暴力法 class Solution: def removeElement(self, nums: List[int], val: int) -> int: i, l = 0, len(nums) From 934dd4e3131485802edd5d7c45fd60dd4f8f2827 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 15:19:23 -0500 Subject: [PATCH 10/76] =?UTF-8?q?Update=200977.=E6=9C=89=E5=BA=8F=E6=95=B0?= =?UTF-8?q?=E7=BB=84=E7=9A=84=E5=B9=B3=E6=96=B9.md?= 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 | 41 ++++++++++++++++++-------- 1 file changed, 28 insertions(+), 13 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 57f8de02..4fbdd1cd 100644 --- a/problems/0977.有序数组的平方.md +++ b/problems/0977.有序数组的平方.md @@ -140,22 +140,37 @@ class Solution { Python： ```Python +（版本一）双指针法 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 + l, r, i = 0, len(nums)-1, len(nums)-1 + res = [float('inf')] * len(nums) # 需要提前定义列表，存放结果 + while l <= r: + if nums[l] ** 2 < nums[r] ** 2: # 左右边界进行对比，找出最大值 + res[i] = nums[r] ** 2 + r -= 1 # 右指针往左移动 else: - ans[k] = rm - j -= 1 - k -= 1 - return ans + res[i] = nums[l] ** 2 + l += 1 # 左指针往右移动 + i -= 1 # 存放结果的指针需要往前平移一位 + return res +``` + +```Python +（版本二）暴力排序法 +class Solution: + def sortedSquares(self, nums: List[int]) -> List[int]: + for i in range(len(nums)): + nums[i] *= nums[i] + nums.sort() + return nums +``` + +```Python +（版本三）暴力排序法+列表推导法 +class Solution: + def sortedSquares(self, nums: List[int]) -> List[int]: + return sorted(x*x for x in nums) ``` Go： From 5f65e3ba24617dc81b199bcbb356f47517f7fbee Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 15:32:36 -0500 Subject: [PATCH 11/76] =?UTF-8?q?Update=200209.=E9=95=BF=E5=BA=A6=E6=9C=80?= =?UTF-8?q?=E5=B0=8F=E7=9A=84=E5=AD=90=E6=95=B0=E7=BB=84.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0209.长度最小的子数组.md | 46 ++++++++++++++++++----- 1 file changed, 36 insertions(+), 10 deletions(-) diff --git a/problems/0209.长度最小的子数组.md b/problems/0209.长度最小的子数组.md index 5a8f91af..d7ae4780 100644 --- a/problems/0209.长度最小的子数组.md +++ b/problems/0209.长度最小的子数组.md @@ -173,18 +173,44 @@ class Solution { Python： ```python +（版本一）滑动窗口法 class Solution: def minSubArrayLen(self, s: int, nums: List[int]) -> int: - res = float("inf") # 定义一个无限大的数 - Sum = 0 # 滑动窗口数值之和 - i = 0 # 滑动窗口起始位置 - for j in range(len(nums)): - Sum += nums[j] - while Sum >= s: - res = min(res, j-i+1) - Sum -= nums[i] - i += 1 - return 0 if res == float("inf") else res + l = len(nums) + left = 0 + right = 0 + min_len = float('inf') + cur_sum = 0 #当前的累加值 + + while right < l: + cur_sum += nums[right] + + while cur_sum >= s: # 当前累加值大于目标值 + min_len = min(min_len, right - left + 1) + cur_sum -= nums[left] + left += 1 + + right += 1 + + return min_len if min_len != float('inf') else 0 +``` + +```python +（版本二）暴力法 +class Solution: + def minSubArrayLen(self, s: int, nums: List[int]) -> int: + l = len(nums) + min_len = float('inf') + + for i in range(l): + cur_sum = 0 + for j in range(i, l): + cur_sum += nums[j] + if cur_sum >= s: + min_len = min(min_len, j - i + 1) + break + + return min_len if min_len != float('inf') else 0 ``` Go： From f45b1f1d28a668e2c90ad66ac0d2141d59fc73ae Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 16:19:13 -0500 Subject: [PATCH 12/76] =?UTF-8?q?Update=200203.=E7=A7=BB=E9=99=A4=E9=93=BE?= =?UTF-8?q?=E8=A1=A8=E5=85=83=E7=B4=A0.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0203.移除链表元素.md | 20 +++++++++++++------- 1 file changed, 13 insertions(+), 7 deletions(-) diff --git a/problems/0203.移除链表元素.md b/problems/0203.移除链表元素.md index b52f16ea..6a0de282 100644 --- a/problems/0203.移除链表元素.md +++ b/problems/0203.移除链表元素.md @@ -307,21 +307,27 @@ public ListNode removeElements(ListNode head, int val) { Python： ```python +（版本一）虚拟头节点法 # 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: - if cur.next.val == val: - cur.next = cur.next.next #删除cur.next节点 + def removeElements(self, head: Optional[ListNode], val: int) -> Optional[ListNode]: + # 创建虚拟头部节点以简化删除过程 + dummy_head = ListNode(next = head) + + # 遍历列表并删除值为val的节点 + current = dummy_head + while current.next: + if current.next.val == val: + current.next = current.next.next else: - cur = cur.next + current = current.next + return dummy_head.next + ``` Go： From 70d8379ff7fe99edbbd2d006b9b3d9a5068c66fc Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 16:26:16 -0500 Subject: [PATCH 13/76] =?UTF-8?q?Update=200707.=E8=AE=BE=E8=AE=A1=E9=93=BE?= =?UTF-8?q?=E8=A1=A8.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0707.设计链表.md | 294 +++++++++++++++++----------------- 1 file changed, 144 insertions(+), 150 deletions(-) diff --git a/problems/0707.设计链表.md b/problems/0707.设计链表.md index c72b7327..aa04d0e1 100644 --- a/problems/0707.设计链表.md +++ b/problems/0707.设计链表.md @@ -485,172 +485,166 @@ class MyLinkedList { Python： ```python -# 单链表 -class MyLinkedList1: - - def __init__(self): - self.dummy_head = Node()# 添加虚拟头指针，便于操作 - self.size = 0 # 设置一个链表长度的属性，便于后续操作，注意每次增和删的时候都要更新 - - def get(self, index): - """ - :type index: int - :rtype: int - """ - if index < 0 or index >= self.size: - return -1 - cur = self.dummy_head.next - while(index): - cur = cur.next - index -= 1 - return cur.val - - def addAtHead(self, val): - """ - :type val: int - :rtype: None - """ - new_node = Node(val) - new_node.next = self.dummy_head.next - self.dummy_head.next = new_node - self.size += 1 - - def addAtTail(self, val): - """ - :type val: int - :rtype: None - """ - new_node = Node(val) - cur = self.dummy_head - while(cur.next): - cur = cur.next - cur.next = new_node - self.size += 1 - - def addAtIndex(self, index, val): - """ - :type index: int - :type val: int - :rtype: None - """ - if index < 0: - self.addAtHead(val) - return - elif index == self.size: - self.addAtTail(val) - return - elif index > self.size: - return - - node = Node(val) - cur = self.dummy_head - while(index): - cur = cur.next - index -= 1 - node.next = cur.next - cur.next = node - self.size += 1 - - def deleteAtIndex(self, index): - """ - :type index: int - :rtype: None - """ - if index < 0 or index >= self.size: - return - pre = self.dummy_head - while(index): - pre = pre.next - index -= 1 - pre.next = pre.next.next - self.size -= 1 - -# 双链表 -# 相对于单链表, Node新增了prev属性 -class Node: - - def __init__(self, val=0, next = None, prev = None): +（版本一）单链表法 +class ListNode: + def __init__(self, val=0, next=None): self.val = val self.next = next - self.prev = prev - + class MyLinkedList: - def __init__(self): - self._head, self._tail = Node(0), Node(0) # 虚拟节点 - self._head.next, self._tail.prev = self._tail, self._head - self._count = 0 # 添加的节点数 - - def _get_node(self, index: int) -> Node: - # 当index小于_count//2时, 使用_head查找更快, 反之_tail更快 - if index >= self._count // 2: - # 使用prev往前找 - node = self._tail - for _ in range(self._count - index): - node = node.prev - else: - # 使用next往后找 - node = self._head - for _ in range(index + 1): - node = node.next - return node - - - def _update(self, prev: Node, next: Node, val: int) -> None: - """ - 更新节点 - :param prev: 相对于更新的前一个节点 - :param next: 相对于更新的后一个节点 - :param val: 要添加的节点值 - """ - # 计数累加 - self._count += 1 - node = Node(val) - prev.next, next.prev = node, node - node.prev, node.next = prev, next + self.dummy_head = ListNode() + self.size = 0 def get(self, index: int) -> int: - """ - Get the value of the index-th node in the linked list. If the index is invalid, return -1. - """ - if 0 <= index < self._count: - node = self._get_node(index) - return node.val - else: + if index < 0 or index >= self.size: return -1 + + current = self.dummy_head.next + for i in range(index): + current = current.next + + return current.val def addAtHead(self, val: int) -> None: - """ - Add a node of value val before the first element of the linked list. After the insertion, the new node will be the first node of the linked list. - """ - self._update(self._head, self._head.next, val) + self.dummy_head.next = ListNode(val, self.dummy_head.next) + self.size += 1 def addAtTail(self, val: int) -> None: - """ - Append a node of value val to the last element of the linked list. - """ - self._update(self._tail.prev, self._tail, val) + current = self.dummy_head + while current.next: + current = current.next + current.next = ListNode(val) + self.size += 1 def addAtIndex(self, index: int, val: int) -> None: - """ - Add a node of value val before the index-th node in the linked list. If index equals to the length of linked list, the node will be appended to the end of linked list. If index is greater than the length, the node will not be inserted. - """ - if index < 0: - index = 0 - elif index > self._count: + if index < 0 or index > self.size: return - node = self._get_node(index) - self._update(node.prev, node, val) + + current = self.dummy_head + for i in range(index): + current = current.next + current.next = ListNode(val, current.next) + self.size += 1 def deleteAtIndex(self, index: int) -> None: - """ - Delete the index-th node in the linked list, if the index is valid. - """ - if 0 <= index < self._count: - node = self._get_node(index) - # 计数-1 - self._count -= 1 - node.prev.next, node.next.prev = node.next, node.prev + if index < 0 or index >= self.size: + return + + current = self.dummy_head + for i in range(index): + current = current.next + current.next = current.next.next + self.size -= 1 + + +# Your MyLinkedList object will be instantiated and called as such: +# obj = MyLinkedList() +# param_1 = obj.get(index) +# obj.addAtHead(val) +# obj.addAtTail(val) +# obj.addAtIndex(index,val) +# obj.deleteAtIndex(index) +``` + + +```python +（版本二）双链表法 +class ListNode: + def __init__(self, val=0, prev=None, next=None): + self.val = val + self.prev = prev + self.next = next + +class MyLinkedList: + def __init__(self): + self.head = None + self.tail = None + self.size = 0 + + def get(self, index: int) -> int: + if index < 0 or index >= self.size: + return -1 + + if index < self.size // 2: + current = self.head + for i in range(index): + current = current.next + else: + current = self.tail + for i in range(self.size - index - 1): + current = current.prev + + return current.val + + def addAtHead(self, val: int) -> None: + new_node = ListNode(val, None, self.head) + if self.head: + self.head.prev = new_node + else: + self.tail = new_node + self.head = new_node + self.size += 1 + + def addAtTail(self, val: int) -> None: + new_node = ListNode(val, self.tail, None) + if self.tail: + self.tail.next = new_node + else: + self.head = new_node + self.tail = new_node + self.size += 1 + + def addAtIndex(self, index: int, val: int) -> None: + if index < 0 or index > self.size: + return + + if index == 0: + self.addAtHead(val) + elif index == self.size: + self.addAtTail(val) + else: + if index < self.size // 2: + current = self.head + for i in range(index - 1): + current = current.next + else: + current = self.tail + for i in range(self.size - index): + current = current.prev + new_node = ListNode(val, current, current.next) + current.next.prev = new_node + current.next = new_node + self.size += 1 + + def deleteAtIndex(self, index: int) -> None: + if index < 0 or index >= self.size: + return + + if index == 0: + self.head = self.head.next + if self.head: + self.head.prev = None + else: + self.tail = None + elif index == self.size - 1: + self.tail = self.tail.prev + if self.tail: + self.tail.next = None + else: + self.head = None + else: + if index < self.size // 2: + current = self.head + for i in range(index): + current = current.next + else: + current = self.tail + for i in range(self.size - index - 1): + current = current.prev + current.prev.next = current.next + current.next.prev = current.prev + self.size -= 1 From 55ea26c5bdac70997c6ad9833448cdf3c1a7e0d1 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 17:18:48 -0500 Subject: [PATCH 14/76] =?UTF-8?q?Update=200206.=E7=BF=BB=E8=BD=AC=E9=93=BE?= =?UTF-8?q?=E8=A1=A8.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0206.翻转链表.md | 40 ++++++++++------------------------- 1 file changed, 11 insertions(+), 29 deletions(-) diff --git a/problems/0206.翻转链表.md b/problems/0206.翻转链表.md index d558e783..0425e182 100644 --- a/problems/0206.翻转链表.md +++ b/problems/0206.翻转链表.md @@ -193,9 +193,9 @@ class Solution { } ``` -Python迭代法： +Python ```python -#双指针 +（版本一）双指针法 # Definition for singly-linked list. # class ListNode: # def __init__(self, val=0, next=None): @@ -205,7 +205,7 @@ class Solution: def reverseList(self, head: ListNode) -> ListNode: cur = head pre = None - while(cur!=None): + while cur: 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 @@ -217,6 +217,7 @@ class Solution: Python递归法： ```python +（版本二）递归法 # Definition for singly-linked list. # class ListNode: # def __init__(self, val=0, next=None): @@ -224,36 +225,17 @@ Python递归法： # self.next = next class Solution: def reverseList(self, head: ListNode) -> ListNode: - - def reverse(pre,cur): - if not cur: - return pre - - tmp = cur.next - cur.next = pre - - return reverse(cur,tmp) - - return reverse(None,head) + return self.reverse(head, None) + def reverse(self, cur: ListNode, pre: ListNode) -> ListNode: + if cur == None: + return pre + temp = cur.next + cur.next = pre + return self.reverse(temp, cur) ``` -Python递归法从后向前： -```python -# 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: Optional[ListNode]) -> Optional[ListNode]: - if not head or not head.next: return head - p = self.reverseList(head.next) - head.next.next = head - head.next = None - return p -``` Go： From d92104209fecd89626df38414d03c78371f251e5 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 17:20:34 -0500 Subject: [PATCH 15/76] =?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 --- .../0024.两两交换链表中的节点.md | 23 +++++++++---------- 1 file changed, 11 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 4dc051aa..2c171dde 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 @@ -186,21 +186,20 @@ Python： class Solution: def swapPairs(self, head: ListNode) -> ListNode: - res = ListNode(next=head) - pre = res + dummy_head = ListNode(next=head) + current = dummy_head - # 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 + # 必须有cur的下一个和下下个才能交换，否则说明已经交换结束了 + while current.next and current.next.next: + temp = current.next # 防止节点修改 + temp1 = current.next.next.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 + current.next = current.next.next + current.next.next = temp + temp.next = temp1 + current = current.next.next + return dummy_head.next - pre = pre.next.next - return res.next ``` Go： From 190477400ebf0842343588c4e9e15c5eab412eb4 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 17:22:46 -0500 Subject: [PATCH 16/76] =?UTF-8?q?Update=200019.=E5=88=A0=E9=99=A4=E9=93=BE?= =?UTF-8?q?=E8=A1=A8=E7=9A=84=E5=80=92=E6=95=B0=E7=AC=ACN=E4=B8=AA?= =?UTF-8?q?=E8=8A=82=E7=82=B9.md?= 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 | 28 ++++++++++++------- 1 file changed, 18 insertions(+), 10 deletions(-) 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 a11ff8ba..c6f5bfc7 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 @@ -127,21 +127,29 @@ Python: # 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+1步 + # 创建一个虚拟节点，并将其下一个指针设置为链表的头部 + dummy_head = ListNode(0, head) + + # 创建两个指针，慢指针和快指针，并将它们初始化为虚拟节点 + slow = fast = dummy_head + + # 快指针比慢指针快 n+1 步 + for i in range(n+1): fast = fast.next - n -= 1 - while(fast!=None): + + # 移动两个指针，直到快速指针到达链表的末尾 + while fast: 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 + + # 通过更新第 (n-1) 个节点的 next 指针删除第 n 个节点 + slow.next = slow.next.next + + return dummy_head.next + ``` Go: ```Go From e4072d9a00b12c6de6949091378bdd8d95b5296c Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 17:25:36 -0500 Subject: [PATCH 17/76] =?UTF-8?q?Update=20=E9=9D=A2=E8=AF=95=E9=A2=9802.07?= =?UTF-8?q?.=E9=93=BE=E8=A1=A8=E7=9B=B8=E4=BA=A4.md?= 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 | 63 ++++++++++++++++++++++++- 1 file changed, 62 insertions(+), 1 deletion(-) 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 30f5c467..4bcbd1f9 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 @@ -152,7 +152,7 @@ public class Solution { ### Python ```python - +（版本一）求长度，同时出发 class Solution: def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode: lenA, lenB = 0, 0 @@ -178,7 +178,68 @@ class Solution: curB = curB.next return None ``` +```python +（版本二）求长度，同时出发 （代码复用） +class Solution: + def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode: + lenA = self.getLength(headA) + lenB = self.getLength(headB) + + # 通过移动较长的链表，使两链表长度相等 + if lenA > lenB: + headA = self.moveForward(headA, lenA - lenB) + else: + headB = self.moveForward(headB, lenB - lenA) + + # 将两个头向前移动，直到它们相交 + while headA and headB: + if headA == headB: + return headA + headA = headA.next + headB = headB.next + + return None + + def getLength(self, head: ListNode) -> int: + length = 0 + while head: + length += 1 + head = head.next + return length + + def moveForward(self, head: ListNode, steps: int) -> ListNode: + while steps > 0: + head = head.next + steps -= 1 + return head +``` +```python +（版本三）等比例法 +# 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: + # 处理边缘情况 + if not headA or not headB: + return None + + # 在每个链表的头部初始化两个指针 + pointerA = headA + pointerB = headB + + # 遍历两个链表直到指针相交 + while pointerA != pointerB: + # 将指针向前移动一个节点 + pointerA = pointerA.next if pointerA else headB + pointerB = pointerB.next if pointerB else headA + + # 如果相交，指针将位于交点节点，如果没有交点，值为None + return pointerA +``` ### Go ```go From b2bfb8016642f388389250df57bad7e9cce82a3f Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 17:27:27 -0500 Subject: [PATCH 18/76] =?UTF-8?q?Update=200142.=E7=8E=AF=E5=BD=A2=E9=93=BE?= =?UTF-8?q?=E8=A1=A8II.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0142.环形链表II.md | 50 ++++++++++++++++++++++++++------- 1 file changed, 40 insertions(+), 10 deletions(-) diff --git a/problems/0142.环形链表II.md b/problems/0142.环形链表II.md index 46df4777..e80a715a 100644 --- a/problems/0142.环形链表II.md +++ b/problems/0142.环形链表II.md @@ -221,25 +221,55 @@ public class Solution { Python： ```python +（版本一）快慢指针法 +# Definition for singly-linked list. +# class ListNode: +# def __init__(self, x): +# self.val = x +# self.next = None + + class Solution: def detectCycle(self, head: ListNode) -> ListNode: - slow, fast = head, head + slow = head + fast = head + while fast and fast.next: slow = slow.next fast = fast.next.next - # 如果相遇 + + # If there is a cycle, the slow and fast pointers will eventually meet if slow == fast: - p = head - q = slow - while p!=q: - p = p.next - q = q.next - #你也可以return q - return p - + # Move one of the pointers back to the start of the list + slow = head + while slow != fast: + slow = slow.next + fast = fast.next + return slow + # If there is no cycle, return None return None ``` +```python +（版本二）集合法 +# Definition for singly-linked list. +# class ListNode: +# def __init__(self, x): +# self.val = x +# self.next = None + +class Solution: + def detectCycle(self, head: ListNode) -> ListNode: + visited = set() + + while head: + if head in visited: + return head + visited.add(head) + head = head.next + + return None +``` Go： ```go From dd58553088514af1a1a019fb9e6d4b2e41b0aeb5 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 19:26:39 -0500 Subject: [PATCH 19/76] =?UTF-8?q?Update=20=E4=BA=8C=E5=8F=89=E6=A0=91?= =?UTF-8?q?=E7=9A=84=E9=80=92=E5=BD=92=E9=81=8D=E5=8E=86.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/二叉树的递归遍历.md | 53 +++++++++++++--------------- 1 file changed, 24 insertions(+), 29 deletions(-) diff --git a/problems/二叉树的递归遍历.md b/problems/二叉树的递归遍历.md index 5a9a670a..8d5a5985 100644 --- a/problems/二叉树的递归遍历.md +++ b/problems/二叉树的递归遍历.md @@ -174,50 +174,45 @@ class Solution { Python： ```python # 前序遍历-递归-LC144_二叉树的前序遍历 +# 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 preorderTraversal(self, root: TreeNode) -> List[int]: - # 保存结果 - result = [] - - def traversal(root: TreeNode): - if root == None: - return - result.append(root.val) # 前序 - traversal(root.left) # 左 - traversal(root.right) # 右 + if not root: + return [] + + left = self.preorderTraversal(root.left) + right = self.preorderTraversal(root.right) + + return [root.val] + left + right - traversal(root) - return result # 中序遍历-递归-LC94_二叉树的中序遍历 class Solution: def inorderTraversal(self, root: TreeNode) -> List[int]: - result = [] + if root is None: + return [] - def traversal(root: TreeNode): - if root == None: - return - traversal(root.left) # 左 - result.append(root.val) # 中序 - traversal(root.right) # 右 + left = self.inorderTraversal(root.left) + right = self.inorderTraversal(root.right) - traversal(root) - return result + return left + [root.val] + right # 后序遍历-递归-LC145_二叉树的后序遍历 class Solution: def postorderTraversal(self, root: TreeNode) -> List[int]: - result = [] + if not root: + return [] - def traversal(root: TreeNode): - if root == None: - return - traversal(root.left) # 左 - traversal(root.right) # 右 - result.append(root.val) # 后序 + left = self.postorderTraversal(root.left) + right = self.postorderTraversal(root.right) - traversal(root) - return result + return left + right + [root.val] ``` Go： From dd1da7fc548c7ab7c1e0964f0ef50410be53e551 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 19:27:58 -0500 Subject: [PATCH 20/76] =?UTF-8?q?Update=20=E4=BA=8C=E5=8F=89=E6=A0=91?= =?UTF-8?q?=E7=9A=84=E9=80=92=E5=BD=92=E9=81=8D=E5=8E=86.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/二叉树的递归遍历.md | 6 +++++- 1 file changed, 5 insertions(+), 1 deletion(-) diff --git a/problems/二叉树的递归遍历.md b/problems/二叉树的递归遍历.md index 8d5a5985..92f342f0 100644 --- a/problems/二叉树的递归遍历.md +++ b/problems/二叉树的递归遍历.md @@ -191,7 +191,8 @@ class Solution: return [root.val] + left + right - +``` +```python # 中序遍历-递归-LC94_二叉树的中序遍历 class Solution: def inorderTraversal(self, root: TreeNode) -> List[int]: @@ -202,6 +203,9 @@ class Solution: right = self.inorderTraversal(root.right) return left + [root.val] + right +``` +```python + # 后序遍历-递归-LC145_二叉树的后序遍历 class Solution: From ce5b335b7258b7febb2dc831d68bfd3bb6161905 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 19:30:53 -0500 Subject: [PATCH 21/76] =?UTF-8?q?Update=20=E4=BA=8C=E5=8F=89=E6=A0=91?= =?UTF-8?q?=E7=9A=84=E8=BF=AD=E4=BB=A3=E9=81=8D=E5=8E=86.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/二叉树的迭代遍历.md | 8 ++++++-- 1 file changed, 6 insertions(+), 2 deletions(-) diff --git a/problems/二叉树的迭代遍历.md b/problems/二叉树的迭代遍历.md index 35cf4077..8b241465 100644 --- a/problems/二叉树的迭代遍历.md +++ b/problems/二叉树的迭代遍历.md @@ -258,7 +258,9 @@ class Solution: if node.left: stack.append(node.left) return result - +``` +```python + # 中序遍历-迭代-LC94_二叉树的中序遍历 class Solution: def inorderTraversal(self, root: TreeNode) -> List[int]: @@ -279,7 +281,9 @@ class Solution: # 取栈顶元素右结点 cur = cur.right return result - + ``` + ```python + # 后序遍历-迭代-LC145_二叉树的后序遍历 class Solution: def postorderTraversal(self, root: TreeNode) -> List[int]: From 0521f762d90b1af39fa4f134d219e2d208988e3b Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 19:48:57 -0500 Subject: [PATCH 22/76] =?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 | 3 +-- 1 file changed, 1 insertion(+), 2 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 e777c2a4..29deee11 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 @@ -178,12 +178,11 @@ python3代码： # self.val = val # self.left = left # self.right = right -from collections import deque class Solution: def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]: if not root: return [] - queue = deque([root]) + queue = collections.deque([root]) result = [] while queue: level = [] From 42f85c8a8f8c637dbb484ca0d0c9ab88973c3a0b Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 19:56:26 -0500 Subject: [PATCH 23/76] =?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 | 36 ++++++++++++----------- 1 file changed, 19 insertions(+), 17 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 29deee11..a4164b2c 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 @@ -512,27 +512,29 @@ python代码： class Solution: """二叉树层序遍历II迭代解法""" +# 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]]: - results = [] if not root: - return results - - from collections import deque - que = deque([root]) - - while que: - result = [] - for _ in range(len(que)): - cur = que.popleft() - result.append(cur.val) + return [] + queue = collections.deque([root]) + result = [] + while queue: + level = [] + for _ in range(len(queue)): + cur = queue.popleft() + level.append(cur.val) if cur.left: - que.append(cur.left) + queue.append(cur.left) if cur.right: - que.append(cur.right) - results.append(result) - - results.reverse() - return results + queue.append(cur.right) + result.append(level) + return result[::-1] ``` Java： From 52092d0153828a14179b41d041e99121a9b4f1f2 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 20:01:07 -0500 Subject: [PATCH 24/76] =?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 | 46 +++++++++++------------ 1 file changed, 23 insertions(+), 23 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 a4164b2c..fdfde822 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 @@ -835,35 +835,35 @@ 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() + + queue = collections.deque([root]) + right_view = [] + + while queue: + level_size = len(queue) + + for i in range(level_size): + node = queue.popleft() + + if i == level_size - 1: + right_view.append(node.val) + if node.left: - quene.append(node.left) + queue.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 + queue.append(node.right) + + return right_view ``` From 95616fa9a8a147f3255ead5523af0a4de51844b9 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 20:03:09 -0500 Subject: [PATCH 25/76] =?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 | 47 ++++++++++++++--------- 1 file changed, 29 insertions(+), 18 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 fdfde822..f4a5c466 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 @@ -1121,27 +1121,38 @@ python代码： class Solution: """二叉树层平均值迭代解法""" +# 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]: - results = [] if not root: - return results + return [] - from collections import deque - que = deque([root]) - - while que: - size = len(que) - sum_ = 0 - for _ in range(size): - cur = que.popleft() - sum_ += cur.val - if cur.left: - que.append(cur.left) - if cur.right: - que.append(cur.right) - results.append(sum_ / size) - - return results + queue = collections.deque([root]) + averages = [] + + while queue: + size = len(queue) + level_sum = 0 + + for i in range(size): + node = queue.popleft() + + + level_sum += node.val + + if node.left: + queue.append(node.left) + if node.right: + queue.append(node.right) + + averages.append(level_sum / size) + + return averages ``` java: From 7be18e2dd372a17288c8cb215b52dc5a06c1a240 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 20:04:54 -0500 Subject: [PATCH 26/76] =?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 | 40 ++++++++++++++--------- 1 file changed, 24 insertions(+), 16 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 f4a5c466..ed2c683d 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 @@ -1426,28 +1426,36 @@ 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: - """N叉树的层序遍历迭代法""" - def levelOrder(self, root: 'Node') -> List[List[int]]: - results = [] if not root: - return results + return [] - from collections import deque - que = deque([root]) + result = [] + queue = collections.deque([root]) - while que: - result = [] - for _ in range(len(que)): - cur = que.popleft() - result.append(cur.val) - # cur.children 是 Node 对象组成的列表，也可能为 None - if cur.children: - que.extend(cur.children) - results.append(result) + while queue: + level_size = len(queue) + level = [] - return results + for _ in range(level_size): + node = queue.popleft() + level.append(node.val) + + for child in node.children: + queue.append(child) + + result.append(level) + + return result ``` ```python From 994db141aa831866a5551d7c01c5d676252b9876 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 20:06:39 -0500 Subject: [PATCH 27/76] =?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 | 39 ++++++++++++++++------- 1 file changed, 27 insertions(+), 12 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 ed2c683d..6235ad21 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 @@ -1761,22 +1761,37 @@ 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 largestValues(self, root: TreeNode) -> List[int]: - if root is None: + if not root: return [] - queue = [root] - out_list = [] + + result = [] + queue = collections.deque([root]) + while queue: - length = len(queue) - in_list = [] - for _ in range(length): - curnode = queue.pop(0) - in_list.append(curnode.val) - if curnode.left: queue.append(curnode.left) - if curnode.right: queue.append(curnode.right) - out_list.append(max(in_list)) - return out_list + level_size = len(queue) + max_val = float('-inf') + + for _ in range(level_size): + node = queue.popleft() + max_val = max(max_val, node.val) + + if node.left: + queue.append(node.left) + + if node.right: + queue.append(node.right) + + result.append(max_val) + + return result ``` java代码： From ffab57f2c330f35a624bd7c9ce7dc9d444f0ff79 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 20:09:12 -0500 Subject: [PATCH 28/76] =?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 | 48 ++++++++++++----------- 1 file changed, 26 insertions(+), 22 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 6235ad21..47969b25 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 @@ -2096,36 +2096,40 @@ class Solution { python代码： ```python -# 层序遍历解法 +""" +# Definition for a Node. +class Node: + def __init__(self, val: int = 0, left: 'Node' = None, right: 'Node' = None, next: 'Node' = None): + self.val = val + self.left = left + self.right = right + self.next = next +""" class Solution: def connect(self, root: 'Node') -> 'Node': if not root: - return None - queue = [root] + return root + + queue = collections.deque([root]) + while queue: - n = len(queue) - for i in range(n): - node = queue.pop(0) + level_size = len(queue) + prev = None + + for i in range(level_size): + node = queue.popleft() + + if prev: + prev.next = node + + prev = node + if node.left: queue.append(node.left) + if node.right: queue.append(node.right) - if i == n - 1: - break - node.next = queue[0] - return root - -# 链表解法 -class Solution: - def connect(self, root: 'Node') -> 'Node': - first = root - while first: - cur = first - while cur: # 遍历每一层的节点 - if cur.left: cur.left.next = cur.right # 找左节点的next - if cur.right and cur.next: cur.right.next = cur.next.left # 找右节点的next - cur = cur.next # cur同层移动到下一节点 - first = first.left # 从本层扩展到下一层 + return root ``` From 65085c4e9bcf8c478fe67caad7dd0dceaa5911e0 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 20:10:51 -0500 Subject: [PATCH 29/76] =?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 | 44 ++++++++++++++++------- 1 file changed, 32 insertions(+), 12 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 47969b25..89e7a126 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 @@ -2381,21 +2381,41 @@ python代码： ```python # 层序遍历解法 +""" +# Definition for a Node. +class Node: + def __init__(self, val: int = 0, left: 'Node' = None, right: 'Node' = None, next: 'Node' = None): + self.val = val + self.left = left + self.right = right + self.next = next +""" + class Solution: def connect(self, root: 'Node') -> 'Node': if not root: - return None - queue = [root] - while queue: # 遍历每一层 - length = len(queue) - tail = None # 每一层维护一个尾节点 - for i in range(length): # 遍历当前层 - curnode = queue.pop(0) - if tail: - tail.next = curnode # 让尾节点指向当前节点 - tail = curnode # 让当前节点成为尾节点 - if curnode.left : queue.append(curnode.left) - if curnode.right: queue.append(curnode.right) + return root + + queue = collections.deque([root]) + + while queue: + level_size = len(queue) + prev = None + + for i in range(level_size): + node = queue.popleft() + + if prev: + prev.next = node + + prev = node + + if node.left: + queue.append(node.left) + + if node.right: + queue.append(node.right) + return root ``` From 61173277185bf7948ffb576212cc31fd383778e3 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 20:12:16 -0500 Subject: [PATCH 30/76] =?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 | 31 ++++++++++++++--------- 1 file changed, 19 insertions(+), 12 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 89e7a126..13694207 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 @@ -2664,24 +2664,31 @@ class Solution { Python： ```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 maxDepth(self, root: TreeNode) -> int: - if root == None: + if not root: return 0 - - queue_ = [root] + depth = 0 - while queue_: - length = len(queue_) - 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) + queue = collections.deque([root]) + + while queue: depth += 1 - + for _ in range(len(queue)): + node = queue.popleft() + if node.left: + queue.append(node.left) + if node.right: + queue.append(node.right) + return depth + ``` Go： From de7c67c35a5d3d0e423072d0b7cb02876c628786 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 20:13:38 -0500 Subject: [PATCH 31/76] =?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 | 33 ++++++++++++----------- 1 file changed, 18 insertions(+), 15 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 13694207..c2ad9508 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 @@ -2938,23 +2938,26 @@ Python 3： # self.right = right class Solution: def minDepth(self, root: TreeNode) -> int: - if root == None: + if not root: return 0 + depth = 0 + queue = collections.deque([root]) + + while queue: + depth += 1 + for _ in range(len(queue)): + node = queue.popleft() + + if not node.left and not node.right: + return depth + + if node.left: + queue.append(node.left) + + if node.right: + queue.append(node.right) - #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 + return depth ``` Go： From 54bcab13e37ee1e609b567411b968bdb5a86abac Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 20:25:30 -0500 Subject: [PATCH 32/76] =?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 | 163 +++++++++++++++++++++++-------- 1 file changed, 120 insertions(+), 43 deletions(-) diff --git a/problems/0226.翻转二叉树.md b/problems/0226.翻转二叉树.md index 16a5be57..63baa409 100644 --- a/problems/0226.翻转二叉树.md +++ b/problems/0226.翻转二叉树.md @@ -314,81 +314,158 @@ class Solution { 递归法：前序遍历： ```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 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 + root.left, root.right = root.right, root.left + self.invertTree(root.left) + self.invertTree(root.right) return root ``` -递归法：后序遍历： +迭代法：前序遍历： ```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 invertTree(self, root: TreeNode) -> TreeNode: - if root is None: + if not root: + return None + stack = [root] + while stack: + node = stack.pop() + node.left, node.right = node.right, node.left + if node.left: + stack.append(node.left) + if node.right: + stack.append(node.right) + return root +``` + + +递归法：中序遍历： +```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 invertTree(self, root: TreeNode) -> TreeNode: + if not root: + return None + self.invertTree(root.left) + root.left, root.right = root.right, root.left + self.invertTree(root.left) + return root +``` + +迭代法：中序遍历： +```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 invertTree(self, root: TreeNode) -> TreeNode: + if not root: + return None + stack = [root] + while stack: + node = stack.pop() + if node.left: + stack.append(node.left) + node.left, node.right = node.right, node.left + if node.left: + stack.append(node.left) + return root +``` + + +递归法：后序遍历： +```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 invertTree(self, root: TreeNode) -> TreeNode: + if not root: return None self.invertTree(root.left) self.invertTree(root.right) root.left, root.right = root.right, root.left - return root + return root ``` -迭代法：深度优先遍历（前序遍历）： +迭代法：后序遍历： ```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 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 + return None + stack = [root] + while stack: + node = stack.pop() if node.left: - st.append(node.left) #左 + stack.append(node.left) + if node.right: + stack.append(node.right) + node.left, node.right = node.right, node.left + return root ``` + + + + 迭代法：广度优先遍历（层序遍历）： ```python -import collections +# 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 invertTree(self, root: TreeNode) -> TreeNode: - queue = collections.deque() #使用deque() - if root: - queue.append(root) + if not root: + return None + + queue = collections.deque([root]) while queue: - size = len(queue) - for i in range(size): + for i in range(len(queue)): 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 -``` -迭代法：广度优先遍历（层序遍历），和之前的层序遍历写法一致： -```python -class Solution: - def invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]: - if not root: return root - from collections import deque - que=deque([root]) - while que: - size=len(que) - for i in range(size): - cur=que.popleft() - cur.left, cur.right = cur.right, cur.left - if cur.left: que.append(cur.left) - if cur.right: que.append(cur.right) + node.left, node.right = node.right, node.left + if node.left: queue.append(node.left) + if node.right: queue.append(node.right) return root + ``` + ### Go 递归版本的前序遍历 From 918d0ec0faf27870212b11c4820d4ae956ba8da3 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 20:40:43 -0500 Subject: [PATCH 33/76] =?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 | 40 ++++++++++++++++++-------------- 1 file changed, 23 insertions(+), 17 deletions(-) diff --git a/problems/0101.对称二叉树.md b/problems/0101.对称二叉树.md index b75e9ff2..81ca79a2 100644 --- a/problems/0101.对称二叉树.md +++ b/problems/0101.对称二叉树.md @@ -442,25 +442,31 @@ class Solution: 层次遍历 ```python class Solution: - def isSymmetric(self, root: Optional[TreeNode]) -> bool: + def isSymmetric(self, root: TreeNode) -> bool: if not root: return True - - que = [root] - while que: - this_level_length = len(que) - for i in range(this_level_length // 2): - # 要么其中一个是None但另外一个不是 - if (not que[i] and que[this_level_length - 1 - i]) or (que[i] and not que[this_level_length - 1 - i]): - return False - # 要么两个都不是None - if que[i] and que[i].val != que[this_level_length - 1 - i].val: - return False - for i in range(this_level_length): - if not que[i]: continue - que.append(que[i].left) - que.append(que[i].right) - que = que[this_level_length:] + + queue = collections.deque([root.left, root.right]) + + while queue: + level_size = len(queue) + + if level_size % 2 != 0: + return False + + level_vals = [] + for i in range(level_size): + node = queue.popleft() + if node: + level_vals.append(node.val) + queue.append(node.left) + queue.append(node.right) + else: + level_vals.append(None) + + if level_vals != level_vals[::-1]: + return False + return True ``` From aac7378cb62ed2ec36fc1012f0057844297bda67 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 21:40:49 -0500 Subject: [PATCH 34/76] =?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 | 25 +++++++++++++++-------- 1 file changed, 16 insertions(+), 9 deletions(-) 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 36578fd3..2294c1d9 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 @@ -419,26 +419,33 @@ class solution: return 1 + max(self.maxdepth(root.left), self.maxdepth(root.right)) ``` -迭代法： +层序遍历迭代法： ```python -import collections -class solution: - def maxdepth(self, root: treenode) -> int: +# 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 maxDepth(self, root: TreeNode) -> int: if not root: return 0 - depth = 0 #记录深度 - queue = collections.deque() - queue.append(root) + + depth = 0 + queue = collections.deque([root]) + while queue: - size = len(queue) depth += 1 - for i in range(size): + for _ in range(len(queue)): 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 From 5da51519b5692ef2f450ba38b6299452a00f3b22 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 21:48:04 -0500 Subject: [PATCH 35/76] =?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 | 15 +++++++++------ 1 file changed, 9 insertions(+), 6 deletions(-) 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 2294c1d9..c4d94d4d 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 @@ -452,14 +452,17 @@ class Solution: 递归法： ```python -class solution: - def maxdepth(self, root: 'node') -> int: +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 + + max_depth = 1 + + for child in root.children: + max_depth = max(max_depth, self.maxDepth(child) + 1) + + return max_depth ``` 迭代法： From be27cc547daf4c82a048b71f201c93ed3b2b3e2e Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 21:57:13 -0500 Subject: [PATCH 36/76] =?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 | 33 ++++++++++++++--------- 1 file changed, 21 insertions(+), 12 deletions(-) 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 c4d94d4d..c55ddb3f 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 @@ -467,22 +467,31 @@ class Solution: 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 +""" +# Definition for a Node. +class Node: + def __init__(self, val=None, children=None): + self.val = val + self.children = children +""" + +class Solution: + def maxDepth(self, root: TreeNode) -> int: + if not root: + return 0 + + depth = 0 + queue = collections.deque([root]) + while queue: - size = len(queue) depth += 1 - for i in range(size): + for _ in range(len(queue)): node = queue.popleft() - for j in range(len(node.children)): - if node.children[j]: - queue.append(node.children[j]) + for child in node.children: + queue.append(child) + return depth + ``` 使用栈来模拟后序遍历依然可以 From 1b1b51750db1afcb812ba1250ea4fdc3e0395425 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 22:04:25 -0500 Subject: [PATCH 37/76] =?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 | 48 ++++++++++++----------- 1 file changed, 25 insertions(+), 23 deletions(-) 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 c55ddb3f..b6208169 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 @@ -494,30 +494,32 @@ class Solution: ``` -使用栈来模拟后序遍历依然可以 +使用栈 ```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 +""" +# Definition for a Node. +class Node: + def __init__(self, val=None, children=None): + self.val = val + self.children = children +""" + +class Solution: + def maxDepth(self, root: 'Node') -> int: + if not root: + return 0 + + max_depth = 0 + + stack = [(root, 1)] + + while stack: + node, depth = stack.pop() + max_depth = max(max_depth, depth) + for child in node.children: + stack.append((child, depth + 1)) + + return max_depth ``` From 75d137eb1c0464de0321f611ed66125e322936e7 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Thu, 4 May 2023 00:23:53 -0500 Subject: [PATCH 38/76] =?UTF-8?q?Update=200111.=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.md?= 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 | 59 +++++++++++++++-------- 1 file changed, 39 insertions(+), 20 deletions(-) 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 de36c6f2..bda12ff0 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 @@ -300,44 +300,63 @@ class Solution { 递归法： ```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 minDepth(self, root: TreeNode) -> int: if not root: return 0 + if not root.left and not root.right: return 1 - - min_depth = 10**9 + + left_depth = float('inf') + right_depth = float('inf') + if root.left: - min_depth = min(self.minDepth(root.left), min_depth) # 获得左子树的最小高度 + left_depth = self.minDepth(root.left) if root.right: - min_depth = min(self.minDepth(root.right), min_depth) # 获得右子树的最小高度 - return min_depth + 1 + right_depth = self.minDepth(root.right) + + return 1 + min(left_depth, right_depth) + ``` 迭代法： ```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 minDepth(self, root: TreeNode) -> int: if not root: return 0 - que = deque() - que.append(root) - res = 1 - - while que: - for _ in range(len(que)): - node = que.popleft() - # 当左右孩子都为空的时候，说明是最低点的一层了，退出 + depth = 0 + queue = collections.deque([root]) + + while queue: + depth += 1 + for _ in range(len(queue)): + node = queue.popleft() + if not node.left and not node.right: - return res - if node.left is not None: - que.append(node.left) - if node.right is not None: - que.append(node.right) - res += 1 - return res + return depth + + if node.left: + queue.append(node.left) + + if node.right: + queue.append(node.right) + + return depth ``` From d15c54f43fe0df608d207f269e8bb589c5cebd3d Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Thu, 4 May 2023 00:26:20 -0500 Subject: [PATCH 39/76] =?UTF-8?q?Update=200111.=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.md?= 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 | 33 +++++++++++++++++++++++ 1 file changed, 33 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 bda12ff0..0c086f1b 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 @@ -359,6 +359,39 @@ class Solution: return depth ``` +迭代法： + +```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 minDepth(self, root: TreeNode) -> int: + if not root: + return 0 + + queue = collections.deque([(root, 1)]) + + while queue: + node, depth = queue.popleft() + + # Check if the node is a leaf node + if not node.left and not node.right: + return depth + + # Add left and right child to the queue + if node.left: + queue.append((node.left, depth+1)) + if node.right: + queue.append((node.right, depth+1)) + + return 0 + +``` ## Go From 3d59f732e0479c12657f5a6020b4ed5784222780 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Thu, 4 May 2023 01:35:13 -0500 Subject: [PATCH 40/76] =?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 --- problems/0222.完全二叉树的节点个数.md | 14 ++++++++++++++ 1 file changed, 14 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 d89a9bce..795a6f37 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 @@ -393,6 +393,20 @@ class Solution: # 利用完全二叉树特性 return 2**count-1 return 1+self.countNodes(root.left)+self.countNodes(root.right) ``` +完全二叉树写法3 +```python +class Solution: # 利用完全二叉树特性 + def countNodes(self, root: TreeNode) -> int: + if not root: return 0 + count = 0 + left = root.left; right = root.right + while left and right: + count+=1 + left = left.left; right = right.right + if not left and not right: # 如果同时到底说明是满二叉树，反之则不是 + return (2< Date: Thu, 4 May 2023 21:49:12 -0500 Subject: [PATCH 41/76] =?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 | 18 ++++++++++++++++++ 1 file changed, 18 insertions(+) diff --git a/problems/0110.平衡二叉树.md b/problems/0110.平衡二叉树.md index 804c95eb..a3bc77fb 100644 --- a/problems/0110.平衡二叉树.md +++ b/problems/0110.平衡二叉树.md @@ -532,6 +532,24 @@ class Solution: else: return 1 + max(left_height, right_height) ``` +递归法精简版： + +```python +class Solution: + def isBalanced(self, root: TreeNode) -> bool: + return self.height(root) != -1 + def height(self, node: TreeNode) -> int: + if not node: + return 0 + left = self.height(node.left) + if left == -1: + return -1 + right = self.height(node.right) + if right == -1 or abs(left - right) > 1: + return -1 + return max(left, right) + 1 +``` + 迭代法： From debaa5e4a0cb16a58f226e04b0865ce1b423f122 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Thu, 4 May 2023 22:54:49 -0500 Subject: [PATCH 42/76] =?UTF-8?q?Update=200257.=E4=BA=8C=E5=8F=89=E6=A0=91?= =?UTF-8?q?=E7=9A=84=E6=89=80=E6=9C=89=E8=B7=AF=E5=BE=84.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0257.二叉树的所有路径.md | 65 +++++++++++++++++++++++ 1 file changed, 65 insertions(+) diff --git a/problems/0257.二叉树的所有路径.md b/problems/0257.二叉树的所有路径.md index 8c542cea..c396f4a0 100644 --- a/problems/0257.二叉树的所有路径.md +++ b/problems/0257.二叉树的所有路径.md @@ -468,6 +468,71 @@ class Solution { ``` --- ## Python: + + +递归法+回溯（版本一） +```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 +import copy +from typing import List, Optional + +class Solution: + def binaryTreePaths(self, root: Optional[TreeNode]) -> List[str]: + if not root: + return [] + result = [] + self.generate_paths(root, [], result) + return result + + def generate_paths(self, node: TreeNode, path: List[int], result: List[str]) -> None: + path.append(node.val) + if not node.left and not node.right: + result.append('->'.join(map(str, path))) + if node.left: + self.generate_paths(node.left, copy.copy(path), result) + if node.right: + self.generate_paths(node.right, copy.copy(path), result) + path.pop() + + +``` +递归法+回溯（版本二） +```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 +import copy +from typing import List, Optional + +class Solution: + def binaryTreePaths(self, root: Optional[TreeNode]) -> List[str]: + if not root: + return [] + result = [] + self.generate_paths(root, [], result) + return result + + def generate_paths(self, node: TreeNode, path: List[int], result: List[str]) -> None: + if not node: + return + path.append(node.val) + if not node.left and not node.right: + result.append('->'.join(map(str, path))) + else: + self.generate_paths(node.left, copy.copy(path), result) + self.generate_paths(node.right, copy.copy(path), result) + path.pop() + +``` + 递归法+隐形回溯 ```Python # Definition for a binary tree node. From 3b0000a280b2e6a5b1bb033556cbc77e793736fb Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Thu, 4 May 2023 23:26:06 -0500 Subject: [PATCH 43/76] =?UTF-8?q?Update=200404.=E5=B7=A6=E5=8F=B6=E5=AD=90?= =?UTF-8?q?=E4=B9=8B=E5=92=8C.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0404.左叶子之和.md | 26 +++++++++++++++++--------- 1 file changed, 17 insertions(+), 9 deletions(-) diff --git a/problems/0404.左叶子之和.md b/problems/0404.左叶子之和.md index cf5441c4..617978b7 100644 --- a/problems/0404.左叶子之和.md +++ b/problems/0404.左叶子之和.md @@ -250,19 +250,27 @@ class Solution { **递归后序遍历** ```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 sumOfLeftLeaves(self, root: TreeNode) -> int: - if not root: + def sumOfLeftLeaves(self, root: Optional[TreeNode]) -> int: + if not root: return 0 - left_left_leaves_sum = self.sumOfLeftLeaves(root.left) # 左 - right_left_leaves_sum = self.sumOfLeftLeaves(root.right) # 右 - - cur_left_leaf_val = 0 - if root.left and not root.left.left and not root.left.right: - cur_left_leaf_val = root.left.val + # 检查根节点的左子节点是否为叶节点 + if root.left and not root.left.left and not root.left.right: + left_val = root.left.val + else: + left_val = self.sumOfLeftLeaves(root.left) - return cur_left_leaf_val + left_left_leaves_sum + right_left_leaves_sum # 中 + # 递归地计算右子树左叶节点的和 + right_val = self.sumOfLeftLeaves(root.right) + + return left_val + right_val ``` **迭代** From 4224c17e54e164141ca2f6a08978402ae115fea5 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Fri, 5 May 2023 19:18:26 -0500 Subject: [PATCH 44/76] =?UTF-8?q?Update=20=E9=9D=A2=E8=AF=95=E9=A2=9802.07?= =?UTF-8?q?.=E9=93=BE=E8=A1=A8=E7=9B=B8=E4=BA=A4.md?= 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, 35 insertions(+), 1 deletion(-) 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 4bcbd1f9..dda7f2ad 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 @@ -214,7 +214,41 @@ class Solution: return head ``` ```python -（版本三）等比例法 +（版本三）求长度，同时出发 （代码复用 + 精简） +class Solution: + def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode: + dis = self.getLength(headA) - self.getLength(headB) + + # 通过移动较长的链表，使两链表长度相等 + if dis > 0: + headA = self.moveForward(headA, dis) + else: + headB = self.moveForward(headB, abs(dis)) + + # 将两个头向前移动，直到它们相交 + while headA and headB: + if headA == headB: + return headA + headA = headA.next + headB = headB.next + + return None + + def getLength(self, head: ListNode) -> int: + length = 0 + while head: + length += 1 + head = head.next + return length + + def moveForward(self, head: ListNode, steps: int) -> ListNode: + while steps > 0: + head = head.next + steps -= 1 + return head +``` +```python +（版本四）等比例法 # Definition for singly-linked list. # class ListNode: # def __init__(self, x): From d590d268d717b0241c5d900f70e9356bf3c88903 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Fri, 5 May 2023 19:50:20 -0500 Subject: [PATCH 45/76] =?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 | 55 +++++++++++++++++++---- 1 file changed, 47 insertions(+), 8 deletions(-) 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 1006ea35..0e5683c7 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 @@ -122,6 +122,21 @@ class Solution { ``` Python： +(版本一) 使用数组 +```python +class Solution: + def isAnagram(self, s: str, t: str) -> bool: + record = [0] * 26 + for i in range(len(s)): + record[ord(s[i]) - ord("a")] += 1 + for i in range(len(t)): + record[ord(t[i]) - ord("a")] -= 1 + for i in range(26): + if record[i] != 0: + return False + return True +``` +(版本二) 使用数组 ```python class Solution: def isAnagram(self, s: str, t: str) -> bool: @@ -138,12 +153,13 @@ class Solution: return True ``` -Python写法二（没有使用数组作为哈希表，只是介绍defaultdict这样一种解题思路）： +(版本三) 使用defaultdict ```python +from collections import defaultdict + class Solution: def isAnagram(self, s: str, t: str) -> bool: - from collections import defaultdict s_dict = defaultdict(int) t_dict = defaultdict(int) @@ -156,17 +172,40 @@ class Solution: return s_dict == t_dict ``` -Python写法三(没有使用数组作为哈希表，只是介绍Counter这种更方便的解题思路)： +(版本四) 使用字典 ```python -class Solution(object): +class Solution: def isAnagram(self, s: str, t: str) -> bool: - from collections import Counter - a_count = Counter(s) - b_count = Counter(t) - return a_count == b_count + if len(s) != len(t): + return False + + hash_table_s = {} + hash_table_t = {} + + for i in range(len(s)): + hash_table_s[s[i]] = hash_table_s.get(s[i], 0) + 1 + hash_table_t[t[i]] = hash_table_t.get(t[i], 0) + 1 + + return hash_table_s == hash_table_t ``` +(版本五) 使用排序 + +```python +class Solution: + def isAnagram(self, s: str, t: str) -> bool: + return sorted(s) == sorted(t) +``` +(版本六) 使用Counter + +```python +from collections import Counter + +class Solution: + def isAnagram(self, s: str, t: str) -> bool: + return Counter(s) == Counter(t) +``` Go： ```go From 3065cf38604dd672ad128a445b98c356604eb24e Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Fri, 5 May 2023 20:04:47 -0500 Subject: [PATCH 46/76] =?UTF-8?q?Update=200349.=E4=B8=A4=E4=B8=AA=E6=95=B0?= =?UTF-8?q?=E7=BB=84=E7=9A=84=E4=BA=A4=E9=9B=86.md?= 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 | 34 ++++++++++++++++++-------- 1 file changed, 24 insertions(+), 10 deletions(-) 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 0da0e30c..c2f6ef46 100644 --- a/problems/0349.两个数组的交集.md +++ b/problems/0349.两个数组的交集.md @@ -160,22 +160,29 @@ class Solution { ``` Python3： +（版本一） 使用字典和集合 ```python class Solution: def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]: - val_dict = {} - ans = [] + # 使用哈希表存储一个数组中的所有元素 + table = {} for num in nums1: - val_dict[num] = 1 - - for num in nums2: - if num in val_dict.keys() and val_dict[num] == 1: - ans.append(num) - val_dict[num] = 0 + table[num] = table.get(num, 0) + 1 - return ans + # 使用集合存储结果 + res = set() + for num in nums2: + if num in table: + res.add(num) + del table[num] + + return list(res) +``` +（版本二） 使用数组 + +```python -class Solution: # 使用数组方法 +class Solution: def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]: count1 = [0]*1001 count2 = [0]*1001 @@ -190,7 +197,14 @@ class Solution: # 使用数组方法 return result ``` +（版本三） 使用集合 +```python +class Solution: + def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]: + return list(set(nums1) & set(nums2)) + +``` Go： ```go From eeae3282a87c334f726fc3c8cd5dabd5b71c4612 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Fri, 5 May 2023 20:25:46 -0500 Subject: [PATCH 47/76] =?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 | 96 ++++++++++++++++++++++++++++++-------- 1 file changed, 76 insertions(+), 20 deletions(-) diff --git a/problems/0202.快乐数.md b/problems/0202.快乐数.md index 5ac29e86..7fe8cd8d 100644 --- a/problems/0202.快乐数.md +++ b/problems/0202.快乐数.md @@ -108,23 +108,14 @@ class Solution { ``` Python： +(版本一)使用集合 ```python class Solution: - def isHappy(self, n: int) -> bool: - def calculate_happy(num): - sum_ = 0 - - # 从个位开始依次取，平方求和 - while num: - sum_ += (num % 10) ** 2 - num = num // 10 - return sum_ - - # 记录中间结果 + def isHappy(self, n: int) -> bool: record = set() while True: - n = calculate_happy(n) + n = self.get_sum(n) if n == 1: return True @@ -134,21 +125,86 @@ class Solution: else: record.add(n) -# python的另一种写法 - 通过字符串来计算各位平方和 + def get_sum(self,n: int) -> int: + new_num = 0 + while n: + n, r = divmod(n, 10) + new_num += r ** 2 + return new_num + ``` + (版本二)使用集合 + ```python +class Solution: + def isHappy(self, n: int) -> bool: + record = set() + while n not in record: + record.add(n) + new_num = 0 + n_str = str(n) + for i in n_str: + new_num+=int(i)**2 + if new_num==1: return True + else: n = new_num + return False +``` + (版本三)使用数组 + ```python class Solution: def isHappy(self, n: int) -> bool: record = [] while n not in record: record.append(n) - newn = 0 - nn = str(n) - for i in nn: - newn+=int(i)**2 - if newn==1: return True - n = newn + new_num = 0 + n_str = str(n) + for i in n_str: + new_num+=int(i)**2 + if new_num==1: return True + else: n = new_num return False ``` - + (版本四)使用快慢指针 + ```python +class Solution: + def isHappy(self, n: int) -> bool: + slow = n + fast = n + while self.get_sum(fast) != 1 and self.get_sum(self.get_sum(fast)): + slow = self.get_sum(slow) + fast = self.get_sum(self.get_sum(fast)) + if slow == fast: + return False + return True + def get_sum(self,n: int) -> int: + new_num = 0 + while n: + n, r = divmod(n, 10) + new_num += r ** 2 + return new_num +``` + (版本五)使用集合+精简 + ```python +class Solution: + def isHappy(self, n: int) -> bool: + seen = set() + while n != 1: + n = sum(int(i) ** 2 for i in str(n)) + if n in seen: + return False + seen.add(n) + return True +``` + (版本六)使用数组+精简 + ```python +class Solution: + def isHappy(self, n: int) -> bool: + seen = [] + while n != 1: + n = sum(int(i) ** 2 for i in str(n)) + if n in seen: + return False + seen.append(n) + return True +``` Go： ```go func isHappy(n int) bool { From 7118f36b66e1def134fda97e992affbd0c5cd79b Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Fri, 5 May 2023 20:38:05 -0500 Subject: [PATCH 48/76] =?UTF-8?q?Update=200001.=E4=B8=A4=E6=95=B0=E4=B9=8B?= =?UTF-8?q?=E5=92=8C.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0001.两数之和.md | 49 ++++++++++++++++++++++++++++++++++- 1 file changed, 48 insertions(+), 1 deletion(-) diff --git a/problems/0001.两数之和.md b/problems/0001.两数之和.md index eea3ba7a..1af8787b 100644 --- a/problems/0001.两数之和.md +++ b/problems/0001.两数之和.md @@ -151,7 +151,7 @@ public int[] twoSum(int[] nums, int target) { ``` Python： - +（版本一） 使用字典 ```python class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: @@ -163,6 +163,53 @@ class Solution: records[value] = index # 遍历当前元素，并在map中寻找是否有匹配的key return [] ``` +（版本二）使用集合 +```python +class Solution: + def twoSum(self, nums: List[int], target: int) -> List[int]: + #创建一个集合来存储我们目前看到的数字 + seen = set() + for i, num in enumerate(nums): + complement = target - num + if complement in seen: + return [nums.index(complement), i] + seen.add(num) +``` +（版本三）使用双指针 +```python +class Solution: + def twoSum(self, nums: List[int], target: int) -> List[int]: + # 对输入列表进行排序 + nums_sorted = sorted(nums) + + # 使用双指针 + left = 0 + right = len(nums_sorted) - 1 + while left < right: + current_sum = nums_sorted[left] + nums_sorted[right] + if current_sum == target: + # 如果和等于目标数，则返回两个数的下标 + left_index = nums.index(nums_sorted[left]) + right_index = nums.index(nums_sorted[right]) + if left_index == right_index: + right_index = nums[left_index+1:].index(nums_sorted[right]) + left_index + 1 + return [left_index, right_index] + elif current_sum < target: + # 如果总和小于目标，则将左侧指针向右移动 + left += 1 + else: + # 如果总和大于目标值，则将右指针向左移动 + right -= 1 +``` +（版本四）暴力法 +```python +class Solution: + def twoSum(self, nums: List[int], target: int) -> List[int]: + for i in range(len(nums)): + for j in range(i+1, len(nums)): + if nums[i] + nums[j] == target: + return [i,j] +``` Go： From 0905d2c7871daca4adb18566f86216a83e7894bc Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Fri, 5 May 2023 20:46:57 -0500 Subject: [PATCH 49/76] =?UTF-8?q?Update=200454.=E5=9B=9B=E6=95=B0=E7=9B=B8?= =?UTF-8?q?=E5=8A=A0II.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0454.四数相加II.md | 14 ++++++-------- 1 file changed, 6 insertions(+), 8 deletions(-) diff --git a/problems/0454.四数相加II.md b/problems/0454.四数相加II.md index 411b60e8..7818c02b 100644 --- a/problems/0454.四数相加II.md +++ b/problems/0454.四数相加II.md @@ -125,11 +125,11 @@ class Solution { ``` Python： - +（版本一） 使用字典 ```python class Solution(object): def fourSumCount(self, nums1, nums2, nums3, nums4): - # use a dict to store the elements in nums1 and nums2 and their sum + # 使用字典存储nums1和nums2中的元素及其和 hashmap = dict() for n1 in nums1: for n2 in nums2: @@ -138,7 +138,7 @@ class Solution(object): else: hashmap[n1+n2] = 1 - # if the -(a+b) exists in nums3 and nums4, we shall add the count + # 如果 -(n1+n2) 存在于nums3和nums4, 存入结果 count = 0 for n3 in nums3: for n4 in nums4: @@ -149,20 +149,18 @@ class Solution(object): ``` - +（版本二）使用 defaultdict ```python +from collections import defaultdict class Solution: def fourSumCount(self, nums1: list, nums2: list, nums3: list, nums4: list) -> int: - from collections import defaultdict # You may use normal dict instead. rec, cnt = defaultdict(lambda : 0), 0 - # To store the summary of all the possible combinations of nums1 & nums2, together with their frequencies. for i in nums1: for j in nums2: rec[i+j] += 1 - # To add up the frequencies if the corresponding value occurs in the dictionary for i in nums3: for j in nums4: - cnt += rec.get(-(i+j), 0) # No matched key, return 0. + cnt += rec.get(-(i+j), 0) return cnt ``` From 5cff41ab09e2b3866f3245c3d06d61b3b432f485 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Fri, 5 May 2023 20:53:12 -0500 Subject: [PATCH 50/76] =?UTF-8?q?Update=200454.=E5=9B=9B=E6=95=B0=E7=9B=B8?= =?UTF-8?q?=E5=8A=A0II.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0454.四数相加II.md | 24 +++++++++++++++++++++++- 1 file changed, 23 insertions(+), 1 deletion(-) diff --git a/problems/0454.四数相加II.md b/problems/0454.四数相加II.md index 7818c02b..31f0504f 100644 --- a/problems/0454.四数相加II.md +++ b/problems/0454.四数相加II.md @@ -149,7 +149,29 @@ class Solution(object): ``` -（版本二）使用 defaultdict +（版本二） 使用字典 +```python +class Solution(object): + def fourSumCount(self, nums1, nums2, nums3, nums4): + # 使用字典存储nums1和nums2中的元素及其和 + hashmap = dict() + for n1 in nums1: + for n2 in nums2: + hashmap[n1+n2] = hashmap.get(n1+n2, 0) + 1 + + # 如果 -(n1+n2) 存在于nums3和nums4, 存入结果 + count = 0 + for n3 in nums3: + for n4 in nums4: + key = - n3 - n4 + if key in hashmap: + count += hashmap[key] + return count + + + +``` +（版本三）使用 defaultdict ```python from collections import defaultdict class Solution: From 38bc7021c6f15136e32785398ad17fdab9098d2b Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Fri, 5 May 2023 21:05:34 -0500 Subject: [PATCH 51/76] =?UTF-8?q?Update=200383.=E8=B5=8E=E9=87=91=E4=BF=A1?= =?UTF-8?q?.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0383.赎金信.md | 98 ++++++++++++++------------------------ 1 file changed, 37 insertions(+), 61 deletions(-) diff --git a/problems/0383.赎金信.md b/problems/0383.赎金信.md index d9a184b6..c2d80d8d 100644 --- a/problems/0383.赎金信.md +++ b/problems/0383.赎金信.md @@ -142,34 +142,27 @@ class Solution { ``` -Python写法一（使用数组作为哈希表）： - +（版本一）使用数组 ```python class Solution: def canConstruct(self, ransomNote: str, magazine: str) -> bool: - - arr = [0] * 26 - - for x in magazine: # 记录 magazine里各个字符出现次数 - arr[ord(x) - ord('a')] += 1 - - for x in ransomNote: # 在arr里对应的字符个数做--操作 - if arr[ord(x) - ord('a')] == 0: # 如果没有出现过直接返回 - return False - else: - arr[ord(x) - ord('a')] -= 1 - - return True + ransom_count = [0] * 26 + magazine_count = [0] * 26 + for c in ransomNote: + ransom_count[ord(c) - ord('a')] += 1 + for c in magazine: + magazine_count[ord(c) - ord('a')] += 1 + return all(ransom_count[i] <= magazine_count[i] for i in range(26)) ``` -Python写法二（使用defaultdict）： +（版本二）使用defaultdict ```python +from collections import defaultdict + class Solution: def canConstruct(self, ransomNote: str, magazine: str) -> bool: - from collections import defaultdict - hashmap = defaultdict(int) for x in magazine: @@ -177,59 +170,42 @@ class Solution: for x in ransomNote: value = hashmap.get(x) - if value is None or value == 0: + if not value or not value: return False else: hashmap[x] -= 1 return True ``` - -Python写法三： - -```python -class Solution(object): - def canConstruct(self, ransomNote, magazine): - """ - :type ransomNote: str - :type magazine: str - :rtype: bool - """ - - # use a dict to store the number of letter occurance in ransomNote - hashmap = dict() - for s in ransomNote: - if s in hashmap: - hashmap[s] += 1 - else: - hashmap[s] = 1 - - # check if the letter we need can be found in magazine - for l in magazine: - if l in hashmap: - hashmap[l] -= 1 - - for key in hashmap: - if hashmap[key] > 0: - return False - - return True -``` - -Python写法四： +（版本三）使用字典 ```python class Solution: def canConstruct(self, ransomNote: str, magazine: str) -> bool: - c1 = collections.Counter(ransomNote) - c2 = collections.Counter(magazine) - x = c1 - c2 - #x只保留值大于0的符号，当c1里面的符号个数小于c2时，不会被保留 - #所以x只保留下了，magazine不能表达的 - if(len(x)==0): - return True - else: - return False + counts = {} + for c in magazine: + counts[c] = counts.get(c, 0) + 1 + for c in ransomNote: + if c not in counts or counts[c] == 0: + return False + counts[c] -= 1 + return True +``` +（版本四）使用Counter + +```python +from collections import Counter + +class Solution: + def canConstruct(self, ransomNote: str, magazine: str) -> bool: + return not Counter(ransomNote) - Counter(magazine) +``` +（版本五）使用count + +```python +class Solution: + def canConstruct(self, ransomNote: str, magazine: str) -> bool: + return all(ransomNote.count(c) <= magazine.count(c) for c in set(ransomNote)) ``` Go： From 1bc667eccdb247b9795e28a8793ec51fe9ed4883 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Fri, 5 May 2023 21:06:23 -0500 Subject: [PATCH 52/76] =?UTF-8?q?Update=200383.=E8=B5=8E=E9=87=91=E4=BF=A1?= =?UTF-8?q?.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0383.赎金信.md | 1 + 1 file changed, 1 insertion(+) diff --git a/problems/0383.赎金信.md b/problems/0383.赎金信.md index c2d80d8d..75729437 100644 --- a/problems/0383.赎金信.md +++ b/problems/0383.赎金信.md @@ -206,6 +206,7 @@ class Solution: class Solution: def canConstruct(self, ransomNote: str, magazine: str) -> bool: return all(ransomNote.count(c) <= magazine.count(c) for c in set(ransomNote)) + ``` Go： From 2ddbe7f3404ca4c7e70cf3264b6ba4e02327ca78 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Fri, 5 May 2023 21:30:24 -0500 Subject: [PATCH 53/76] =?UTF-8?q?Update=200015.=E4=B8=89=E6=95=B0=E4=B9=8B?= =?UTF-8?q?=E5=92=8C.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0015.三数之和.md | 95 +++++++++++++++++++---------------- 1 file changed, 53 insertions(+), 42 deletions(-) diff --git a/problems/0015.三数之和.md b/problems/0015.三数之和.md index 26c9eaa2..9cca779b 100644 --- a/problems/0015.三数之和.md +++ b/problems/0015.三数之和.md @@ -298,61 +298,72 @@ class Solution { ``` Python： +（版本一） 双指针 ```Python class Solution: - def threeSum(self, nums): - ans = [] - n = len(nums) + def threeSum(self, nums: List[int]) -> List[List[int]]: + result = [] nums.sort() - # 找出a + b + c = 0 - # a = nums[i], b = nums[left], c = nums[right] - for i in range(n): - left = i + 1 - right = n - 1 - # 排序之后如果第一个元素已经大于零，那么无论如何组合都不可能凑成三元组，直接返回结果就可以了 - if nums[i] > 0: - break - if i >= 1 and nums[i] == nums[i - 1]: # 去重a + + for i in range(len(nums)): + # 如果第一个元素已经大于0，不需要进一步检查 + if nums[i] > 0: + return result + + # 跳过相同的元素以避免重复 + if i > 0 and nums[i] == nums[i - 1]: continue - while left < right: - total = nums[i] + nums[left] + nums[right] - if total > 0: - right -= 1 - elif total < 0: + + left = i + 1 + right = len(nums) - 1 + + while right > left: + sum_ = nums[i] + nums[left] + nums[right] + + if sum_ < 0: left += 1 + elif sum_ > 0: + right -= 1 else: - ans.append([nums[i], nums[left], nums[right]]) - # 去重逻辑应该放在找到一个三元组之后，对b 和 c去重 - while left != right and nums[left] == nums[left + 1]: left += 1 - while left != right and nums[right] == nums[right - 1]: right -= 1 - left += 1 + result.append([nums[i], nums[left], nums[right]]) + + # 跳过相同的元素以避免重复 + while right > left and nums[right] == nums[right - 1]: + right -= 1 + while right > left and nums[left] == nums[left + 1]: + left += 1 + right -= 1 - return ans + left += 1 + + return result ``` -Python (v3): +（版本二） 使用字典 ```python class Solution: def threeSum(self, nums: List[int]) -> List[List[int]]: - if len(nums) < 3: return [] - nums, res = sorted(nums), [] - for i in range(len(nums) - 2): - cur, l, r = nums[i], i + 1, len(nums) - 1 - if res != [] and res[-1][0] == cur: continue # Drop duplicates for the first time. - - while l < r: - if cur + nums[l] + nums[r] == 0: - res.append([cur, nums[l], nums[r]]) - # Drop duplicates for the second time in interation of l & r. Only used when target situation occurs, because that is the reason for dropping duplicates. - while l < r - 1 and nums[l] == nums[l + 1]: - l += 1 - while r > l + 1 and nums[r] == nums[r - 1]: - r -= 1 - if cur + nums[l] + nums[r] > 0: - r -= 1 + result = [] + nums.sort() + # 找出a + b + c = 0 + # a = nums[i], b = nums[j], c = -(a + b) + for i in range(len(nums)): + # 排序之后如果第一个元素已经大于零，那么不可能凑成三元组 + if nums[i] > 0: + break + if i > 0 and nums[i] == nums[i - 1]: #三元组元素a去重 + continue + d = {} + for j in range(i + 1, len(nums)): + if j > i + 2 and nums[j] == nums[j-1] == nums[j-2]: # 三元组元素b去重 + continue + c = 0 - (nums[i] + nums[j]) + if c in d: + result.append([nums[i], nums[j], c]) + d.pop(c) # 三元组元素c去重 else: - l += 1 - return res + d[nums[j]] = j + return result ``` Go： From a5bb942c95f405181ba311e14e1ad601bcfbf435 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Fri, 5 May 2023 21:35:49 -0500 Subject: [PATCH 54/76] =?UTF-8?q?Update=200018.=E5=9B=9B=E6=95=B0=E4=B9=8B?= =?UTF-8?q?=E5=92=8C.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0018.四数之和.md | 53 +++++++++++++++++++---------------- 1 file changed, 29 insertions(+), 24 deletions(-) diff --git a/problems/0018.四数之和.md b/problems/0018.四数之和.md index 5f4c2ec9..8606b2eb 100644 --- a/problems/0018.四数之和.md +++ b/problems/0018.四数之和.md @@ -207,35 +207,40 @@ class Solution { ``` Python： +(版本一) 双指针 ```python -# 双指针法 -class Solution: - def fourSum(self, nums: List[int], target: int) -> List[List[int]]: - + def fourSum(self, nums: List[int], target: int) -> List[List[int]]: + result = [] nums.sort() - n = len(nums) - res = [] - for i in range(n): - if i > 0 and nums[i] == nums[i - 1]: continue # 对nums[i]去重 - for k in range(i+1, n): - if k > i + 1 and nums[k] == nums[k-1]: continue # 对nums[k]去重 - p = k + 1 - q = n - 1 - - while p < q: - if nums[i] + nums[k] + nums[p] + nums[q] > target: q -= 1 - elif nums[i] + nums[k] + nums[p] + nums[q] < target: p += 1 + for k in range(len(nums)): + if nums[k] > target and nums[k] >= 0: + break + if k > 0 and nums[k] == nums[k-1]: + continue + for i in range(k+1, len(nums)): + if nums[k] + nums[i] > target and nums[k] + nums[i] >= 0: + break + if i > k+1 and nums[i] == nums[i-1]: + continue + left, right = i+1, len(nums)-1 + while right > left: + if nums[k] + nums[i] + nums[left] + nums[right] > target: + right -= 1 + elif nums[k] + nums[i] + nums[left] + nums[right] < target: + left += 1 else: - res.append([nums[i], nums[k], nums[p], nums[q]]) - # 对nums[p]和nums[q]去重 - while p < q and nums[p] == nums[p + 1]: p += 1 - while p < q and nums[q] == nums[q - 1]: q -= 1 - p += 1 - q -= 1 - return res + result.append([nums[k], nums[i], nums[left], nums[right]]) + while right > left and nums[right] == nums[right-1]: + right -= 1 + while right > left and nums[left] == nums[left+1]: + left += 1 + right -= 1 + left += 1 + return result ``` +(版本二) 使用字典 + ```python -# 哈希表法 class Solution(object): def fourSum(self, nums, target): """ From dde92b1da3808200925d82f8f523da7b5cd93ca2 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Fri, 5 May 2023 21:36:38 -0500 Subject: [PATCH 55/76] =?UTF-8?q?Update=200018.=E5=9B=9B=E6=95=B0=E4=B9=8B?= =?UTF-8?q?=E5=92=8C.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0018.四数之和.md | 40 +++++++++++++++++++---------------- 1 file changed, 22 insertions(+), 18 deletions(-) diff --git a/problems/0018.四数之和.md b/problems/0018.四数之和.md index 8606b2eb..4fee8358 100644 --- a/problems/0018.四数之和.md +++ b/problems/0018.四数之和.md @@ -209,34 +209,38 @@ class Solution { Python： (版本一) 双指针 ```python - def fourSum(self, nums: List[int], target: int) -> List[List[int]]: - result = [] +class Solution: + def fourSum(self, nums: List[int], target: int) -> List[List[int]]: nums.sort() - for k in range(len(nums)): - if nums[k] > target and nums[k] >= 0: + n = len(nums) + result = [] + for i in range(n): + if nums[i] > target and nums[i] > 0 and target > 0:# 剪枝（可省） break - if k > 0 and nums[k] == nums[k-1]: + if i > 0 and nums[i] == nums[i-1]:# 去重 continue - for i in range(k+1, len(nums)): - if nums[k] + nums[i] > target and nums[k] + nums[i] >= 0: + for j in range(i+1, n): + if nums[i] + nums[j] > target and target > 0: #剪枝（可省） break - if i > k+1 and nums[i] == nums[i-1]: + if j > i+1 and nums[j] == nums[j-1]: # 去重 continue - left, right = i+1, len(nums)-1 - while right > left: - if nums[k] + nums[i] + nums[left] + nums[right] > target: + left, right = j+1, n-1 + while left < right: + s = nums[i] + nums[j] + nums[left] + nums[right] + if s == target: + result.append([nums[i], nums[j], nums[left], nums[right]]) + while left < right and nums[left] == nums[left+1]: + left += 1 + while left < right and nums[right] == nums[right-1]: + right -= 1 + left += 1 right -= 1 - elif nums[k] + nums[i] + nums[left] + nums[right] < target: + elif s < target: left += 1 else: - result.append([nums[k], nums[i], nums[left], nums[right]]) - while right > left and nums[right] == nums[right-1]: - right -= 1 - while right > left and nums[left] == nums[left+1]: - left += 1 right -= 1 - left += 1 return result + ``` (版本二) 使用字典 From c4bc3197502473c5f14be392157bb83ddaa2dbeb Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Fri, 5 May 2023 21:39:52 -0500 Subject: [PATCH 56/76] =?UTF-8?q?Update=200018.=E5=9B=9B=E6=95=B0=E4=B9=8B?= =?UTF-8?q?=E5=92=8C.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0018.四数之和.md | 33 +++++++++++---------------------- 1 file changed, 11 insertions(+), 22 deletions(-) diff --git a/problems/0018.四数之和.md b/problems/0018.四数之和.md index 4fee8358..a4d41d9b 100644 --- a/problems/0018.四数之和.md +++ b/problems/0018.四数之和.md @@ -252,36 +252,25 @@ class Solution(object): :type target: int :rtype: List[List[int]] """ - # use a dict to store value:showtimes - hashmap = dict() - for n in nums: - if n in hashmap: - hashmap[n] += 1 - else: - hashmap[n] = 1 + # 创建一个字典来存储输入列表中每个数字的频率 + freq = {} + for num in nums: + freq[num] = freq.get(num, 0) + 1 - # good thing about using python is you can use set to drop duplicates. + # 创建一个集合来存储最终答案，并遍历4个数字的所有唯一组合 ans = set() - # ans = [] # save results by list() for i in range(len(nums)): for j in range(i + 1, len(nums)): for k in range(j + 1, len(nums)): val = target - (nums[i] + nums[j] + nums[k]) - if val in hashmap: - # make sure no duplicates. + if val in freq: + # 确保没有重复 count = (nums[i] == val) + (nums[j] == val) + (nums[k] == val) - if hashmap[val] > count: - ans_tmp = tuple(sorted([nums[i], nums[j], nums[k], val])) - ans.add(ans_tmp) - # Avoiding duplication in list manner but it cause time complexity increases - # if ans_tmp not in ans: - # ans.append(ans_tmp) - else: - continue - return list(ans) - # if used list() to save results, just - # return ans + if freq[val] > count: + ans.add(tuple(sorted([nums[i], nums[j], nums[k], val]))) + return [list(x) for x in ans] + ``` Go： From b617532ce9b7af1cd163a0ccccb1f956f77e8cd2 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Sat, 6 May 2023 01:55:48 -0500 Subject: [PATCH 57/76] =?UTF-8?q?Update=200513.=E6=89=BE=E6=A0=91=E5=B7=A6?= =?UTF-8?q?=E4=B8=8B=E8=A7=92=E7=9A=84=E5=80=BC.md?= 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 | 102 ++++++++++++++++--------- 1 file changed, 65 insertions(+), 37 deletions(-) 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 5ff25484..26768c74 100644 --- a/problems/0513.找树左下角的值.md +++ b/problems/0513.找树左下角的值.md @@ -271,52 +271,80 @@ class Solution { ### Python - -递归: +（版本一）递归法 + 回溯 ```python class Solution: def findBottomLeftValue(self, root: TreeNode) -> int: - max_depth = -float("INF") - leftmost_val = 0 + self.max_depth = float('-inf') + self.result = None + self.traversal(root, 0) + return self.result + + def traversal(self, node, depth): + if not node.left and not node.right: + if depth > self.max_depth: + self.max_depth = depth + self.result = node.val + return + + if node.left: + depth += 1 + self.traversal(node.left, depth) + depth -= 1 + if node.right: + depth += 1 + self.traversal(node.right, depth) + depth -= 1 - def __traverse(root, cur_depth): - nonlocal max_depth, leftmost_val - if not root.left and not root.right: - if cur_depth > max_depth: - max_depth = cur_depth - leftmost_val = root.val - if root.left: - cur_depth += 1 - __traverse(root.left, cur_depth) - cur_depth -= 1 - if root.right: - cur_depth += 1 - __traverse(root.right, cur_depth) - cur_depth -= 1 - - __traverse(root, 0) - return leftmost_val ``` -迭代 - 层序遍历: +（版本二）递归法+精简 ```python class Solution: def findBottomLeftValue(self, root: TreeNode) -> int: - queue = deque() - if root: - queue.append(root) - result = 0 - while queue: - q_len = len(queue) - for i in range(q_len): - if i == 0: - result = queue[i].val - cur = queue.popleft() - if cur.left: - queue.append(cur.left) - if cur.right: - queue.append(cur.right) - return result + self.max_depth = float('-inf') + self.result = None + self.traversal(root, 0) + return self.result + + def traversal(self, node, depth): + if not node.left and not node.right: + if depth > self.max_depth: + self.max_depth = depth + self.result = node.val + return + + if node.left: + self.traversal(node.left, depth+1) + if node.right: + self.traversal(node.right, depth+1) +``` + +(版本三） 迭代法 +```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 +from collections import deque +class Solution: + def findBottomLeftValue(self, root: Optional[TreeNode]) -> int: + queue = deque([root]) + while queue: + size = len(queue) + leftmost = queue[0].val + for i in range(size): + node = queue.popleft() + if node.left: + queue.append(node.left) + if node.right: + queue.append(node.right) + if not queue: + return leftmost + + ``` ### Go From bb8b112ae0079dd0d09d6a075e17888616ba3a55 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Sat, 6 May 2023 16:51:15 -0500 Subject: [PATCH 58/76] =?UTF-8?q?Update=200344.=E5=8F=8D=E8=BD=AC=E5=AD=97?= =?UTF-8?q?=E7=AC=A6=E4=B8=B2.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0344.反转字符串.md | 58 +++++++++++++++++++++++++++++++- 1 file changed, 57 insertions(+), 1 deletion(-) diff --git a/problems/0344.反转字符串.md b/problems/0344.反转字符串.md index 775cfc58..03c82767 100644 --- a/problems/0344.反转字符串.md +++ b/problems/0344.反转字符串.md @@ -174,6 +174,7 @@ class Solution { ``` Python： +（版本一） 双指针 ```python class Solution: def reverseString(self, s: List[str]) -> None: @@ -190,7 +191,62 @@ class Solution: right -= 1 ``` - +（版本二） 使用栈 +```python +class Solution: + def reverseString(self, s: List[str]) -> None: + """ + Do not return anything, modify s in-place instead. + """ + stack = [] + for char in s: + stack.append(char) + for i in range(len(s)): + s[i] = stack.pop() + +``` +（版本三） 使用range +```python +class Solution: + def reverseString(self, s: List[str]) -> None: + """ + Do not return anything, modify s in-place instead. + """ + n = len(s) + for i in range(n // 2): + s[i], s[n - i - 1] = s[n - i - 1], s[i] + +``` +（版本四） 使用reversed +```python +class Solution: + def reverseString(self, s: List[str]) -> None: + """ + Do not return anything, modify s in-place instead. + """ + s[:] = reversed(s) + +``` +（版本五） 使用切片 +```python +class Solution: + def reverseString(self, s: List[str]) -> None: + """ + Do not return anything, modify s in-place instead. + """ + s[:] = s[::-1] + +``` +（版本六） 使用列表推导 +```python +class Solution: + def reverseString(self, s: List[str]) -> None: + """ + Do not return anything, modify s in-place instead. + """ + s[:] = [s[i] for i in range(len(s) - 1, -1, -1)] + +``` Go： ```Go func reverseString(s []byte) { From a37ae94618edd6104f9d0815bc0d97e185175f1e Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Sat, 6 May 2023 16:54:34 -0500 Subject: [PATCH 59/76] =?UTF-8?q?Update=200344.=E5=8F=8D=E8=BD=AC=E5=AD=97?= =?UTF-8?q?=E7=AC=A6=E4=B8=B2.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0344.反转字符串.md | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/problems/0344.反转字符串.md b/problems/0344.反转字符串.md index 03c82767..1c74f9aa 100644 --- a/problems/0344.反转字符串.md +++ b/problems/0344.反转字符串.md @@ -183,8 +183,8 @@ class Solution: """ left, right = 0, len(s) - 1 - # 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 + # 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(len(s)//2)更低 + # 因为while每次循环需要进行条件判断，而range函数不需要，直接生成数字，因此时间复杂度更低。推荐使用range while left < right: s[left], s[right] = s[right], s[left] left += 1 From abcedd41a87e6923cfd39705765cdd5716b31ad0 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Sat, 6 May 2023 16:59:12 -0500 Subject: [PATCH 60/76] =?UTF-8?q?Update=20=E5=89=91=E6=8C=87Offer05.?= =?UTF-8?q?=E6=9B=BF=E6=8D=A2=E7=A9=BA=E6=A0=BC.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/剑指Offer05.替换空格.md | 1 + 1 file changed, 1 insertion(+) diff --git a/problems/剑指Offer05.替换空格.md b/problems/剑指Offer05.替换空格.md index 2e1ee1de..833d4628 100644 --- a/problems/剑指Offer05.替换空格.md +++ b/problems/剑指Offer05.替换空格.md @@ -266,6 +266,7 @@ func replaceSpace(s string) string { python： +####因为字符串是不可变类型，所以操作字符串需要将其转换为列表，因此空间复杂度不可能不为O(1) ```python class Solution: def replaceSpace(self, s: str) -> str: From 3958cc166a1ab090f0060e45bab41535170cd186 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Sat, 6 May 2023 16:59:48 -0500 Subject: [PATCH 61/76] =?UTF-8?q?Update=20=E5=89=91=E6=8C=87Offer05.?= =?UTF-8?q?=E6=9B=BF=E6=8D=A2=E7=A9=BA=E6=A0=BC.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/剑指Offer05.替换空格.md | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/problems/剑指Offer05.替换空格.md b/problems/剑指Offer05.替换空格.md index 833d4628..d930702f 100644 --- a/problems/剑指Offer05.替换空格.md +++ b/problems/剑指Offer05.替换空格.md @@ -265,8 +265,8 @@ func replaceSpace(s string) string { -python： -####因为字符串是不可变类型，所以操作字符串需要将其转换为列表，因此空间复杂度不可能不为O(1) +#python： +#因为字符串是不可变类型，所以操作字符串需要将其转换为列表，因此空间复杂度不可能不为O(1) ```python class Solution: def replaceSpace(self, s: str) -> str: From 799325bacebd60e3088e5ab4a82ef599ba1d3e47 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Sat, 6 May 2023 17:00:54 -0500 Subject: [PATCH 62/76] =?UTF-8?q?Update=20=E5=89=91=E6=8C=87Offer05.?= =?UTF-8?q?=E6=9B=BF=E6=8D=A2=E7=A9=BA=E6=A0=BC.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/剑指Offer05.替换空格.md | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/problems/剑指Offer05.替换空格.md b/problems/剑指Offer05.替换空格.md index d930702f..a6c32d30 100644 --- a/problems/剑指Offer05.替换空格.md +++ b/problems/剑指Offer05.替换空格.md @@ -265,8 +265,8 @@ func replaceSpace(s string) string { -#python： -#因为字符串是不可变类型，所以操作字符串需要将其转换为列表，因此空间复杂度不可能不为O(1) +# python： +# 因为字符串是不可变类型，所以操作字符串需要将其转换为列表，因此空间复杂度不可能不为O(1) ```python class Solution: def replaceSpace(self, s: str) -> str: From dc7ac6337d5ae0a462cc332a5a1a68de43095212 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Sat, 6 May 2023 17:01:39 -0500 Subject: [PATCH 63/76] =?UTF-8?q?Update=20=E5=89=91=E6=8C=87Offer05.?= =?UTF-8?q?=E6=9B=BF=E6=8D=A2=E7=A9=BA=E6=A0=BC.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/剑指Offer05.替换空格.md | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/problems/剑指Offer05.替换空格.md b/problems/剑指Offer05.替换空格.md index a6c32d30..d86cc6dc 100644 --- a/problems/剑指Offer05.替换空格.md +++ b/problems/剑指Offer05.替换空格.md @@ -265,8 +265,8 @@ func replaceSpace(s string) string { -# python： -# 因为字符串是不可变类型，所以操作字符串需要将其转换为列表，因此空间复杂度不可能不为O(1) +python： +#### 因为字符串是不可变类型，所以操作字符串需要将其转换为列表，因此空间复杂度不可能不为O(1) ```python class Solution: def replaceSpace(self, s: str) -> str: From c09089accda7c3cc46bc283eb63edfc9563894ab Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Sat, 6 May 2023 17:10:18 -0500 Subject: [PATCH 64/76] =?UTF-8?q?Update=20=E5=89=91=E6=8C=87Offer05.?= =?UTF-8?q?=E6=9B=BF=E6=8D=A2=E7=A9=BA=E6=A0=BC.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/剑指Offer05.替换空格.md | 34 ++++++++++++++++++++------ 1 file changed, 27 insertions(+), 7 deletions(-) diff --git a/problems/剑指Offer05.替换空格.md b/problems/剑指Offer05.替换空格.md index d86cc6dc..dbad781e 100644 --- a/problems/剑指Offer05.替换空格.md +++ b/problems/剑指Offer05.替换空格.md @@ -266,7 +266,8 @@ func replaceSpace(s string) string { python： -#### 因为字符串是不可变类型，所以操作字符串需要将其转换为列表，因此空间复杂度不可能不为O(1) +#### 因为字符串是不可变类型，所以操作字符串需要将其转换为列表，因此空间复杂度不可能为O(1) +（版本一）转换成列表，并且添加相匹配的空间，然后进行填充 ```python class Solution: def replaceSpace(self, s: str) -> str: @@ -291,14 +292,22 @@ class Solution: return ''.join(res) ``` - +（版本二）添加空列表，添加匹配的结果 +```python +class Solution: + def replaceSpace(self, s: str) -> str: + res = [] + for i in range(len(s)): + if s[i] == ' ': + res.append('%20') + else: + res.append(s[i]) + return ''.join(res) +``` +（版本三）使用切片 ```python class Solution: def replaceSpace(self, s: str) -> str: - # method 1 - Very rude - return "%20".join(s.split(" ")) - - # method 2 - Reverse the s when counting in for loop, then update from the end. n = len(s) for e, i in enumerate(s[::-1]): print(i, e) @@ -307,7 +316,18 @@ class Solution: print("") return s ``` - +（版本四）使用join + split +```python +class Solution: + def replaceSpace(self, s: str) -> str: + return "%20".join(s.split(" ")) +``` +（版本五）使用replace +```python +class Solution: + def replaceSpace(self, s: str) -> str: + return s.replace(' ', '%20') +``` javaScript: ```js From c75140d2572be2030e64a8aa18a7cc336b7a5a74 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Sat, 6 May 2023 17:26:08 -0500 Subject: [PATCH 65/76] =?UTF-8?q?Update=200151.=E7=BF=BB=E8=BD=AC=E5=AD=97?= =?UTF-8?q?=E7=AC=A6=E4=B8=B2=E9=87=8C=E7=9A=84=E5=8D=95=E8=AF=8D.md?= 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 | 136 +++---------------- 1 file changed, 21 insertions(+), 115 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 8fa7c77c..a3232264 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 @@ -434,134 +434,40 @@ class Solution { ``` python: - +（版本一）先删除空白，然后整个反转，最后单词反转。 +### 因为字符串是不可变类型，所以反转单词的时候，需要将其转换成列表，然后通过join函数再将其转换成列表，所以空间复杂度不是O(1) ```Python class Solution: - #1.去除多余的空格 - def trim_spaces(self, s): - n = len(s) - left = 0 - right = n-1 - - while left <= right and s[left] == ' ': #去除开头的空格 - left += 1 - while left <= right and s[right] == ' ': #去除结尾的空格 - right -= 1 - tmp = [] - while left <= right: #去除单词中间多余的空格 - if s[left] != ' ': - tmp.append(s[left]) - elif tmp[-1] != ' ': #当前位置是空格，但是相邻的上一个位置不是空格，则该空格是合理的 - tmp.append(s[left]) - left += 1 - return tmp - - #2.翻转字符数组 - def reverse_string(self, nums, left, right): - while left < right: - nums[left], nums[right] = nums[right], nums[left] - left += 1 - right -= 1 - return None - - #3.翻转每个单词 - def reverse_each_word(self, nums): - start = 0 - end = 0 - n = len(nums) - while start < n: - while end < n and nums[end] != ' ': - end += 1 - self.reverse_string(nums, start, end-1) - start = end + 1 - end += 1 - return None - -#4.翻转字符串里的单词 - def reverseWords(self, s): #测试用例："the sky is blue" - l = self.trim_spaces(s) #输出：['t', 'h', 'e', ' ', 's', 'k', 'y', ' ', 'i', 's', ' ', 'b', 'l', 'u', 'e' - self.reverse_string(l, 0, len(l)-1) #输出：['e', 'u', 'l', 'b', ' ', 's', 'i', ' ', 'y', 'k', 's', ' ', 'e', 'h', 't'] - self.reverse_each_word(l) #输出：['b', 'l', 'u', 'e', ' ', 'i', 's', ' ', 's', 'k', 'y', ' ', 't', 'h', 'e'] - return ''.join(l) #输出：blue is sky the - + def reverseWords(self, s: str) -> str: + # 删除前后空白 + s = s.strip() + # 反转整个字符串 + s = s[::-1] + # 将字符串拆分为单词，并反转每个单词 + s = ' '.join(word[::-1] for word in s.split()) + return s ``` +（版本二）使用双指针 ```python class Solution: def reverseWords(self, s: str) -> str: - # method 1 - Rude but work & efficient method. - s_list = [i for i in s.split(" ") if len(i) > 0] - return " ".join(s_list[::-1]) + # 将字符串拆分为单词，即转换成列表类型 + words = s.split() - # method 2 - Carlo's idea - def trim_head_tail_space(ss: str): - p = 0 - while p < len(ss) and ss[p] == " ": - p += 1 - return ss[p:] + # 反转单词 + left, right = 0, len(words) - 1 + while left < right: + words[left], words[right] = words[right], words[left] + left += 1 + right -= 1 - # Trim the head and tail space - s = trim_head_tail_space(s) - s = trim_head_tail_space(s[::-1])[::-1] - - pf, ps, s = 0, 0, s[::-1] # Reverse the string. - while pf < len(s): - if s[pf] == " ": - # Will not excede. Because we have clean the tail space. - if s[pf] == s[pf + 1]: - s = s[:pf] + s[pf + 1:] - continue - else: - s = s[:ps] + s[ps: pf][::-1] + s[pf:] - ps, pf = pf + 1, pf + 2 - else: - pf += 1 - return s[:ps] + s[ps:][::-1] # Must do the last step, because the last word is omit though the pointers are on the correct positions, + # 将列表转换成字符串 + return " ".join(words) ``` -```python -class Solution: # 使用双指针法移除空格 - def reverseWords(self, s: str) -> str: - - def removeextraspace(s): - start = 0; end = len(s)-1 - while s[start]==' ': - start+=1 - while s[end]==' ': - end-=1 - news = list(s[start:end+1]) - slow = fast = 0 - while fast0 and news[fast]==news[fast-1]==' ': - fast+=1 - news[slow]=news[fast] - slow+=1; fast+=1 - #return "".join(news[:slow]) - return news[:slow] - def reversestr(s): - left,right = 0,len(s)-1 - news = list(s) - while left Date: Sat, 6 May 2023 17:26:56 -0500 Subject: [PATCH 66/76] =?UTF-8?q?Update=200151.=E7=BF=BB=E8=BD=AC=E5=AD=97?= =?UTF-8?q?=E7=AC=A6=E4=B8=B2=E9=87=8C=E7=9A=84=E5=8D=95=E8=AF=8D.md?= 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 | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) 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 a3232264..062ccca5 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 @@ -435,7 +435,7 @@ class Solution { python: （版本一）先删除空白，然后整个反转，最后单词反转。 -### 因为字符串是不可变类型，所以反转单词的时候，需要将其转换成列表，然后通过join函数再将其转换成列表，所以空间复杂度不是O(1) +#### 因为字符串是不可变类型，所以反转单词的时候，需要将其转换成列表，然后通过join函数再将其转换成列表，所以空间复杂度不是O(1) ```Python class Solution: def reverseWords(self, s: str) -> str: From 7cfd38940249adb1f659da29b69edc2ae3083113 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Sat, 6 May 2023 17:52:08 -0500 Subject: [PATCH 67/76] =?UTF-8?q?Update=20=E5=89=91=E6=8C=87Offer58-II.?= =?UTF-8?q?=E5=B7=A6=E6=97=8B=E8=BD=AC=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 --- .../剑指Offer58-II.左旋转字符串.md | 54 ++++++++++--------- 1 file changed, 28 insertions(+), 26 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 f368514f..6cd88456 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 @@ -142,15 +142,16 @@ class Solution { ``` python: +（版本一）使用切片 ```python -# 方法一：可以使用切片方法 class Solution: def reverseLeftWords(self, s: str, n: int) -> str: - return s[n:] + s[0:n] + return s[n:] + s[:n] ``` +（版本二）使用reversed + join + ```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) @@ -161,32 +162,29 @@ class Solution: return "".join(s) ``` +（版本三）自定义reversed函数 ```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: - def reverse_sub(lst, left, right): - while left < right: - lst[left], lst[right] = lst[right], lst[left] - left += 1 - right -= 1 + def reverseLeftWords(self, s: str, n: int) -> str: + s_list = list(s) - 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) + self.reverse(s_list, 0, n - 1) + self.reverse(s_list, n, len(s_list) - 1) + self.reverse(s_list, 0, len(s_list) - 1) -# 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 + return ''.join(s_list) + + def reverse(self, s, start, end): + while start < end: + s[start], s[end] = s[end], s[start] + start += 1 + end -= 1 ``` +（版本四）使用 模 +下标 ```python 3 -#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 = '' @@ -196,17 +194,21 @@ class Solution: return new_s ``` +（版本五）使用 模 + 切片 ```python 3 -# 方法五：另类的切片方法 class Solution: def reverseLeftWords(self, s: str, n: int) -> str: - n = len(s) - s = s + s - return s[k : n+k] + l = len(s) + # 复制输入字符串与它自己连接 + s = s + s + + # 计算旋转字符串的起始索引 + k = n % (l * 2) + + # 从连接的字符串中提取旋转后的字符串并返回 + return s[k : k + l] -# 时间复杂度：O(n) -# 空间复杂度：O(n) ``` Go： From 3cf55028680832e76c82f9f819443b79339f1d49 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Sat, 6 May 2023 19:39:34 -0500 Subject: [PATCH 68/76] =?UTF-8?q?Update=200028.=E5=AE=9E=E7=8E=B0strStr.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0028.实现strStr.md | 99 +++++++++++++++++++++-------------- 1 file changed, 59 insertions(+), 40 deletions(-) diff --git a/problems/0028.实现strStr.md b/problems/0028.实现strStr.md index 263c1689..5cc04994 100644 --- a/problems/0028.实现strStr.md +++ b/problems/0028.实现strStr.md @@ -692,23 +692,38 @@ class Solution { ``` Python3： +（版本一）前缀表（不减一） + ```python -//暴力解法： -class Solution(object): - def strStr(self, haystack, needle): - """ - :type haystack: str - :type needle: str - :rtype: int - """ - m, n = len(haystack), len(needle) - for i in range(m): - if haystack[i:i+n] == needle: - return i - return -1 +class Solution: + def strStr(self, haystack: str, needle: str) -> int: + if len(needle) == 0: + return 0 + next = self.getNext(needle) + j = 0 + for i in range(len(haystack)): + while j >= 1 and haystack[i] != needle[j]: + j = next[j-1] + if haystack[i] == needle[j]: + j += 1 + if j == len(needle): + return i - len(needle) + 1 + return -1 + + def getNext(self, needle): + next = [0] * len(needle) + j = 0 + next[0] = j + for i in range(1, len(needle)): + while j >= 1 and needle[i] != needle[j]: + j = next[j-1] + if needle[i] == needle[j]: + j += 1 + next[i] = j + return next ``` +（版本二）前缀表（减一） ```python -// 方法一 class Solution: def strStr(self, haystack: str, needle: str) -> int: a = len(needle) @@ -739,8 +754,11 @@ class Solution: return next ``` + +（版本三）前缀表（减一） + + ```python -// 方法二 class Solution: def strStr(self, haystack: str, needle: str) -> int: a = len(needle) @@ -774,35 +792,36 @@ class Solution: return next ``` +（版本四）暴力法 +```python +class Solution(object): + def strStr(self, haystack, needle): + """ + :type haystack: str + :type needle: str + :rtype: int + """ + m, n = len(haystack), len(needle) + for i in range(m): + if haystack[i:i+n] == needle: + return i + return -1 +``` +（版本五）使用 index ```python -// 前缀表（不减一）Python实现 class Solution: def strStr(self, haystack: str, needle: str) -> int: - if len(needle) == 0: - return 0 - next = self.getNext(needle) - j = 0 - for i in range(len(haystack)): - while j >= 1 and haystack[i] != needle[j]: - j = next[j-1] - if haystack[i] == needle[j]: - j += 1 - if j == len(needle): - return i - len(needle) + 1 - return -1 - - def getNext(self, needle): - next = [0] * len(needle) - j = 0 - next[0] = j - for i in range(1, len(needle)): - while j >= 1 and needle[i] != needle[j]: - j = next[j-1] - if needle[i] == needle[j]: - j += 1 - next[i] = j - return next + try: + return haystack.index(needle) + except ValueError: + return -1 ``` +（版本六）使用 find +```python +class Solution: + def strStr(self, haystack: str, needle: str) -> int: + return haystack.find(needle) + Go： From 1fbd900c6a942b86d9c1d2f394ed89023e22a50d Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Sat, 6 May 2023 19:47:55 -0500 Subject: [PATCH 69/76] =?UTF-8?q?Update=200028.=E5=AE=9E=E7=8E=B0strStr.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0028.实现strStr.md | 133 ++++++++++++---------------------- 1 file changed, 46 insertions(+), 87 deletions(-) diff --git a/problems/0028.实现strStr.md b/problems/0028.实现strStr.md index 5cc04994..ff13db39 100644 --- a/problems/0028.实现strStr.md +++ b/problems/0028.实现strStr.md @@ -692,107 +692,66 @@ class Solution { ``` Python3： -（版本一）前缀表（不减一） +（版本一）前缀表（减一） +```python +class Solution: + def getNext(self, next, s): + j = -1 + next[0] = j + for i in range(1, len(s)): + while j >= 0 and s[i] != s[j+1]: + j = next[j] + if s[i] == s[j+1]: + j += 1 + next[i] = j + + def strStr(self, haystack: str, needle: str) -> int: + if not needle: + return 0 + next = [0] * len(needle) + self.getNext(next, needle) + j = -1 + for i in range(len(haystack)): + while j >= 0 and haystack[i] != needle[j+1]: + j = next[j] + if haystack[i] == needle[j+1]: + j += 1 + if j == len(needle) - 1: + return i - len(needle) + 1 + return -1 +``` +（版本二）前缀表（不减一） ```python class Solution: + def getNext(self, next: List[int], s: str) -> None: + j = 0 + next[0] = 0 + for i in range(1, len(s)): + while j > 0 and s[i] != s[j]: + j = next[j - 1] + if s[i] == s[j]: + j += 1 + next[i] = j + def strStr(self, haystack: str, needle: str) -> int: if len(needle) == 0: return 0 - next = self.getNext(needle) + next = [0] * len(needle) + self.getNext(next, needle) j = 0 for i in range(len(haystack)): - while j >= 1 and haystack[i] != needle[j]: - j = next[j-1] + while j > 0 and haystack[i] != needle[j]: + j = next[j - 1] if haystack[i] == needle[j]: j += 1 if j == len(needle): return i - len(needle) + 1 return -1 - - def getNext(self, needle): - next = [0] * len(needle) - j = 0 - next[0] = j - for i in range(1, len(needle)): - while j >= 1 and needle[i] != needle[j]: - j = next[j-1] - if needle[i] == needle[j]: - j += 1 - next[i] = j - return next -``` -（版本二）前缀表（减一） -```python -class Solution: - def strStr(self, haystack: str, needle: str) -> int: - a = len(needle) - b = len(haystack) - if a == 0: - return 0 - next = self.getnext(a,needle) - p=-1 - for j in range(b): - while p >= 0 and needle[p+1] != haystack[j]: - p = next[p] - if needle[p+1] == haystack[j]: - p += 1 - if p == a-1: - return j-a+1 - return -1 - - def getnext(self,a,needle): - next = ['' for i in range(a)] - k = -1 - next[0] = k - for i in range(1, len(needle)): - while (k > -1 and needle[k+1] != needle[i]): - k = next[k] - if needle[k+1] == needle[i]: - k += 1 - next[i] = k - return next ``` -（版本三）前缀表（减一） - - -```python -class Solution: - def strStr(self, haystack: str, needle: str) -> int: - a = len(needle) - b = len(haystack) - if a == 0: - return 0 - i = j = 0 - next = self.getnext(a, needle) - while(i < b and j < a): - if j == -1 or needle[j] == haystack[i]: - i += 1 - j += 1 - else: - j = next[j] - if j == a: - return i-j - else: - return -1 - - def getnext(self, a, needle): - next = ['' for i in range(a)] - j, k = 0, -1 - next[0] = k - while(j < a-1): - if k == -1 or needle[k] == needle[j]: - k += 1 - j += 1 - next[j] = k - else: - k = next[k] - return next -``` - -（版本四）暴力法 +（版本三）暴力法 ```python class Solution(object): def strStr(self, haystack, needle): @@ -807,7 +766,7 @@ class Solution(object): return i return -1 ``` -（版本五）使用 index +（版本四）使用 index ```python class Solution: def strStr(self, haystack: str, needle: str) -> int: @@ -816,7 +775,7 @@ class Solution: except ValueError: return -1 ``` -（版本六）使用 find +（版本五）使用 find ```python class Solution: def strStr(self, haystack: str, needle: str) -> int: From 995028d7c45099f71d9ef398ad92a8b655e0c4f9 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Sat, 6 May 2023 19:48:45 -0500 Subject: [PATCH 70/76] =?UTF-8?q?Update=200028.=E5=AE=9E=E7=8E=B0strStr.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0028.实现strStr.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/0028.实现strStr.md b/problems/0028.实现strStr.md index ff13db39..b18131f3 100644 --- a/problems/0028.实现strStr.md +++ b/problems/0028.实现strStr.md @@ -780,7 +780,7 @@ class Solution: class Solution: def strStr(self, haystack: str, needle: str) -> int: return haystack.find(needle) - +``` Go： From e8b4db9d7a5cf58e56e58254195cc2a98d926479 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Sat, 6 May 2023 19:49:19 -0500 Subject: [PATCH 71/76] =?UTF-8?q?Update=200028.=E5=AE=9E=E7=8E=B0strStr.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0028.实现strStr.md | 1 + 1 file changed, 1 insertion(+) diff --git a/problems/0028.实现strStr.md b/problems/0028.实现strStr.md index b18131f3..86308b47 100644 --- a/problems/0028.实现strStr.md +++ b/problems/0028.实现strStr.md @@ -780,6 +780,7 @@ class Solution: class Solution: def strStr(self, haystack: str, needle: str) -> int: return haystack.find(needle) + ``` Go： From 5000a978fd41dd32a639935459365fd58e691c37 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Sat, 6 May 2023 20:05:02 -0500 Subject: [PATCH 72/76] =?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 | 39 ++++++++++++++++++++++++-- 1 file changed, 36 insertions(+), 3 deletions(-) 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 98c02a25..5d56ad18 100644 --- a/problems/0459.重复的子字符串.md +++ b/problems/0459.重复的子字符串.md @@ -264,8 +264,7 @@ class Solution { Python： -这里使用了前缀表统一减一的实现方式 - +（版本一） 前缀表 减一 ```python class Solution: def repeatedSubstringPattern(self, s: str) -> bool: @@ -289,7 +288,7 @@ class Solution: return nxt ``` -前缀表（不减一）的代码实现 +（版本二） 前缀表 不减一 ```python class Solution: @@ -314,6 +313,40 @@ class Solution: return nxt ``` + +（版本三） 使用 find + +```python +class Solution: + def repeatedSubstringPattern(self, s: str) -> bool: + n = len(s) + if n <= 1: + return False + ss = s[1:] + s[:-1] + print(ss.find(s)) + return ss.find(s) != -1 +``` + +（版本四） 暴力法 + +```python +class Solution: + def repeatedSubstringPattern(self, s: str) -> bool: + n = len(s) + if n <= 1: + return False + + substr = "" + for i in range(1, n//2 + 1): + if n % i == 0: + substr = s[:i] + if substr * (n//i) == s: + return True + + return False +``` + + Go： 这里使用了前缀表统一减一的实现方式 From f97ff3a1beeb938107bd76dfa38dac3c93430095 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=A8=8B=E5=BA=8F=E5=91=98Carl?= Date: Sun, 7 May 2023 09:28:28 +0800 Subject: [PATCH 73/76] =?UTF-8?q?Update=200151.=E7=BF=BB=E8=BD=AC=E5=AD=97?= =?UTF-8?q?=E7=AC=A6=E4=B8=B2=E9=87=8C=E7=9A=84=E5=8D=95=E8=AF=8D.md?= 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 | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) 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 062ccca5..4474f1c6 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 @@ -435,7 +435,7 @@ class Solution { python: （版本一）先删除空白，然后整个反转，最后单词反转。 -#### 因为字符串是不可变类型，所以反转单词的时候，需要将其转换成列表，然后通过join函数再将其转换成列表，所以空间复杂度不是O(1) +**因为字符串是不可变类型，所以反转单词的时候，需要将其转换成列表，然后通过join函数再将其转换成列表，所以空间复杂度不是O(1)** ```Python class Solution: def reverseWords(self, s: str) -> str: From 3f45b56cef522ca54b648ded97a8d3b0c6a0f919 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=A8=8B=E5=BA=8F=E5=91=98Carl?= Date: Sun, 7 May 2023 09:47:01 +0800 Subject: [PATCH 74/76] =?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 | 68 +---------------------- 1 file changed, 1 insertion(+), 67 deletions(-) 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 0e5683c7..3f4b00dc 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 @@ -122,21 +122,7 @@ class Solution { ``` Python： -(版本一) 使用数组 -```python -class Solution: - def isAnagram(self, s: str, t: str) -> bool: - record = [0] * 26 - for i in range(len(s)): - record[ord(s[i]) - ord("a")] += 1 - for i in range(len(t)): - record[ord(t[i]) - ord("a")] -= 1 - for i in range(26): - if record[i] != 0: - return False - return True -``` -(版本二) 使用数组 + ```python class Solution: def isAnagram(self, s: str, t: str) -> bool: @@ -153,59 +139,7 @@ class Solution: return True ``` -(版本三) 使用defaultdict -```python -from collections import defaultdict - -class Solution: - def isAnagram(self, s: str, t: str) -> bool: - - s_dict = defaultdict(int) - t_dict = defaultdict(int) - - for x in s: - s_dict[x] += 1 - - for x in t: - t_dict[x] += 1 - - return s_dict == t_dict -``` -(版本四) 使用字典 - -```python -class Solution: - def isAnagram(self, s: str, t: str) -> bool: - if len(s) != len(t): - return False - - hash_table_s = {} - hash_table_t = {} - - for i in range(len(s)): - hash_table_s[s[i]] = hash_table_s.get(s[i], 0) + 1 - hash_table_t[t[i]] = hash_table_t.get(t[i], 0) + 1 - - return hash_table_s == hash_table_t -``` - -(版本五) 使用排序 - -```python -class Solution: - def isAnagram(self, s: str, t: str) -> bool: - return sorted(s) == sorted(t) -``` -(版本六) 使用Counter - -```python -from collections import Counter - -class Solution: - def isAnagram(self, s: str, t: str) -> bool: - return Counter(s) == Counter(t) -``` Go： ```go From 6e88b2034d70d261a9cd0388cdcf906e50677195 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=A8=8B=E5=BA=8F=E5=91=98Carl?= Date: Sun, 7 May 2023 09:54:15 +0800 Subject: [PATCH 75/76] =?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 | 24 +++++++++++++++++++++++ 1 file changed, 24 insertions(+) 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 3f4b00dc..09674ecd 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 @@ -138,7 +138,31 @@ class Solution: return False return True ``` +Python写法二（没有使用数组作为哈希表，只是介绍defaultdict这样一种解题思路）： +```python +class Solution: + def isAnagram(self, s: str, t: str) -> bool: + from collections import defaultdict + + s_dict = defaultdict(int) + t_dict = defaultdict(int) + for x in s: + s_dict[x] += 1 + + for x in t: + t_dict[x] += 1 + return s_dict == t_dict +``` +Python写法三(没有使用数组作为哈希表，只是介绍Counter这种更方便的解题思路)： + +```python +class Solution(object): + def isAnagram(self, s: str, t: str) -> bool: + from collections import Counter + a_count = Counter(s) + b_count = Counter(t) + return a_count == b_count Go： From d2abce01594ae1bbd73889040c82b35455b3d93e Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Sat, 6 May 2023 22:54:54 -0500 Subject: [PATCH 76/76] =?UTF-8?q?Update=200112.=E8=B7=AF=E5=BE=84=E6=80=BB?= =?UTF-8?q?=E5=92=8C.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0112.路径总和.md | 267 ++++++++++++++++++++-------------- 1 file changed, 159 insertions(+), 108 deletions(-) diff --git a/problems/0112.路径总和.md b/problems/0112.路径总和.md index 5958de93..9ee2464e 100644 --- a/problems/0112.路径总和.md +++ b/problems/0112.路径总和.md @@ -451,140 +451,191 @@ class Solution { ### 0112.路径总和 -**递归** - +(版本一) 递归 ```python -class solution: - def haspathsum(self, root: treenode, targetsum: int) -> bool: - def isornot(root, targetsum) -> bool: - if (not root.left) and (not root.right) and targetsum == 0: - return true # 遇到叶子节点，并且计数为0 - if (not root.left) and (not root.right): - return false # 遇到叶子节点，计数不为0 - if root.left: - targetsum -= root.left.val # 左节点 - if isornot(root.left, targetsum): return true # 递归，处理左节点 - targetsum += root.left.val # 回溯 - if root.right: - targetsum -= root.right.val # 右节点 - if isornot(root.right, targetsum): return true # 递归，处理右节点 - targetsum += root.right.val # 回溯 - return false - - if root == none: - return false # 别忘记处理空treenode - else: - return isornot(root, targetsum - root.val) +# 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 traversal(self, cur: TreeNode, count: int) -> bool: + if not cur.left and not cur.right and count == 0: # 遇到叶子节点，并且计数为0 + return True + if not cur.left and not cur.right: # 遇到叶子节点直接返回 + return False + + if cur.left: # 左 + count -= cur.left.val + if self.traversal(cur.left, count): # 递归，处理节点 + return True + count += cur.left.val # 回溯，撤销处理结果 -class Solution: # 简洁版 - def hasPathSum(self, root: Optional[TreeNode], targetSum: int) -> bool: - if not root: return False - if root.left==root.right==None and root.val == targetSum: return True - return self.hasPathSum(root.left,targetSum-root.val) or self.hasPathSum(root.right,targetSum-root.val) + if cur.right: # 右 + count -= cur.right.val + if self.traversal(cur.right, count): # 递归，处理节点 + return True + count += cur.right.val # 回溯，撤销处理结果 + + return False + + def hasPathSum(self, root: TreeNode, sum: int) -> bool: + if root is None: + return False + return self.traversal(root, sum - root.val) ``` -**迭代 - 层序遍历** - +(版本二) 递归 + 精简 ```python -class solution: - def haspathsum(self, root: treenode, targetsum: int) -> bool: +# 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 hasPathSum(self, root: TreeNode, sum: int) -> bool: if not root: - return false - - stack = [] # [(当前节点，路径数值), ...] - stack.append((root, root.val)) - - while stack: - cur_node, path_sum = stack.pop() - - if not cur_node.left and not cur_node.right and path_sum == targetsum: - return true - - if cur_node.right: - stack.append((cur_node.right, path_sum + cur_node.right.val)) - - if cur_node.left: - stack.append((cur_node.left, path_sum + cur_node.left.val)) - - return false + return False + if not root.left and not root.right and sum == root.val: + return True + return self.hasPathSum(root.left, sum - root.val) or self.hasPathSum(root.right, sum - root.val) + ``` +(版本三) 迭代 +```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 hasPathSum(self, root: TreeNode, sum: int) -> bool: + if not root: + return False + # 此时栈里要放的是pair<节点指针，路径数值> + st = [(root, root.val)] + while st: + node, path_sum = st.pop() + # 如果该节点是叶子节点了，同时该节点的路径数值等于sum，那么就返回true + if not node.left and not node.right and path_sum == sum: + return True + # 右节点，压进去一个节点的时候，将该节点的路径数值也记录下来 + if node.right: + st.append((node.right, path_sum + node.right.val)) + # 左节点，压进去一个节点的时候，将该节点的路径数值也记录下来 + if node.left: + st.append((node.left, path_sum + node.left.val)) + return False + + + +``` + + ### 0113.路径总和-ii -**递归** - +(版本一) 递归 ```python -class solution: - def pathsum(self, root: treenode, targetsum: int) -> list[list[int]]: +# 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 __init__(self): + self.result = [] + self.path = [] - def traversal(cur_node, remain): - if not cur_node.left and not cur_node.right: - if remain == 0: - result.append(path[:]) - return + def traversal(self, cur, count): + if not cur.left and not cur.right and count == 0: # 遇到了叶子节点且找到了和为sum的路径 + self.result.append(self.path[:]) + return - if cur_node.left: - path.append(cur_node.left.val) - traversal(cur_node.left, remain-cur_node.left.val) - path.pop() + if not cur.left and not cur.right: # 遇到叶子节点而没有找到合适的边，直接返回 + return - if cur_node.right: - path.append(cur_node.right.val) - traversal(cur_node.right, remain-cur_node.right.val) - path.pop() + if cur.left: # 左 （空节点不遍历） + self.path.append(cur.left.val) + count -= cur.left.val + self.traversal(cur.left, count) # 递归 + count += cur.left.val # 回溯 + self.path.pop() # 回溯 - result, path = [], [] + if cur.right: # 右 （空节点不遍历） + self.path.append(cur.right.val) + count -= cur.right.val + self.traversal(cur.right, count) # 递归 + count += cur.right.val # 回溯 + self.path.pop() # 回溯 + + return + + def pathSum(self, root: TreeNode, sum: int) -> List[List[int]]: + self.result.clear() + self.path.clear() if not root: - return [] - path.append(root.val) - traversal(root, targetsum - root.val) - return result + return self.result + self.path.append(root.val) # 把根节点放进路径 + self.traversal(root, sum - root.val) + return self.result ``` -**迭代法，用第二个队列保存目前的总和与路径** - +(版本二) 递归 + 精简 ```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 pathSum(self, root: Optional[TreeNode], targetSum: int) -> List[List[int]]: - if not root: - return [] - que, temp = deque([root]), deque([(root.val, [root.val])]) + def pathSum(self, root: TreeNode, targetSum: int) -> List[List[int]]: + result = [] - while que: - for _ in range(len(que)): - node = que.popleft() - value, path = temp.popleft() - if (not node.left) and (not node.right): - if value == targetSum: - result.append(path) - if node.left: - que.append(node.left) - temp.append((node.left.val+value, path+[node.left.val])) - if node.right: - que.append(node.right) - temp.append((node.right.val+value, path+[node.right.val])) + self.traversal(root, targetSum, [], result) return result + def traversal(self,node, count, path, result): + if not node: + return + path.append(node.val) + count -= node.val + if not node.left and not node.right and count == 0: + result.append(list(path)) + self.traversal(node.left, count, path, result) + self.traversal(node.right, count, path, result) + path.pop() ``` - -**迭代法，前序遍历** - +(版本三) 迭代 ```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 pathSum(self, root: Optional[TreeNode], targetSum: int) -> List[List[int]]: - if not root: return [] - stack, path_stack,result = [[root,root.val]],[[root.val]],[] + def pathSum(self, root: TreeNode, targetSum: int) -> List[List[int]]: + if not root: + return [] + stack = [(root, [root.val])] + res = [] while stack: - cur,cursum = stack.pop() - path = path_stack.pop() - if cur.left==cur.right==None: - if cursum==targetSum: result.append(path) - if cur.right: - stack.append([cur.right,cursum+cur.right.val]) - path_stack.append(path+[cur.right.val]) - if cur.left: - stack.append([cur.left,cursum+cur.left.val]) - path_stack.append(path+[cur.left.val]) - return result + node, path = stack.pop() + if not node.left and not node.right and sum(path) == targetSum: + res.append(path) + if node.right: + stack.append((node.right, path + [node.right.val])) + if node.left: + stack.append((node.left, path + [node.left.val])) + return res + + + ``` ## go