From ee2993e60f2ac22f976f1780ef183f4eeb68a36b Mon Sep 17 00:00:00 2001 From: zhicheng lee <904688436@qq.com> Date: Sat, 9 Apr 2022 11:40:29 +0800 Subject: [PATCH 01/32] =?UTF-8?q?0332.=E9=87=8D=E6=96=B0=E5=AE=89=E6=8E=92?= =?UTF-8?q?=E8=A1=8C=E7=A8=8B=20=E6=B7=BB=E5=8A=A0Java=E4=BB=A3=E7=A0=81?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0332.重新安排行程.md | 37 +++++++++++++++++++++++++++++ 1 file changed, 37 insertions(+) diff --git a/problems/0332.重新安排行程.md b/problems/0332.重新安排行程.md index 041a7f03..ceb64589 100644 --- a/problems/0332.重新安排行程.md +++ b/problems/0332.重新安排行程.md @@ -261,6 +261,43 @@ for (pairtarget : targets[result[result.size() - 1]]) ## 其他语言版本 ### java + +```java +class Solution { + private LinkedList res; + private LinkedList path = new LinkedList<>(); + + public List findItinerary(List> tickets) { + Collections.sort(tickets, (a, b) -> a.get(1).compareTo(b.get(1))); + path.add("JFK"); + boolean[] used = new boolean[tickets.size()]; + backTracking((ArrayList) tickets, used); + return res; + } + + public boolean backTracking(ArrayList> tickets, boolean[] used) { + if (path.size() == tickets.size() + 1) { + res = new LinkedList(path); + return true; + } + + for (int i = 0; i < tickets.size(); i++) { + if (!used[i] && tickets.get(i).get(0).equals(path.getLast())) { + path.add(tickets.get(i).get(1)); + used[i] = true; + + if (backTracking(tickets, used)) { + return true; + } + + used[i] = false; + path.removeLast(); + } + } + return false; + } +} +``` ```java class Solution { From b6633feb6996a46aa0858d73b1803370d5bafc8e Mon Sep 17 00:00:00 2001 From: zhicheng lee <904688436@qq.com> Date: Sun, 10 Apr 2022 10:27:02 +0800 Subject: [PATCH 02/32] =?UTF-8?q?0051.N=E7=9A=87=E5=90=8E=20=E8=B0=83?= =?UTF-8?q?=E6=95=B4Java=E4=BB=A3=E7=A0=81=E5=92=8CPython=E4=BB=A3?= =?UTF-8?q?=E7=A0=81=E7=9A=84=E9=A1=BA=E5=BA=8F=EF=BC=8C=E4=BF=AE=E6=94=B9?= =?UTF-8?q?=E4=B8=BA=E4=B8=8E=E5=85=B6=E5=AE=83=E9=A2=98=E8=A7=A3=E9=A1=BA?= =?UTF-8?q?=E5=BA=8F=E4=B8=80=E8=87=B4?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0051.N皇后.md | 101 ++++++++++++++++++++------------------- 1 file changed, 51 insertions(+), 50 deletions(-) diff --git a/problems/0051.N皇后.md b/problems/0051.N皇后.md index 7eb0d7a0..1b0cf002 100644 --- a/problems/0051.N皇后.md +++ b/problems/0051.N皇后.md @@ -224,56 +224,6 @@ public: ## 其他语言补充 - -### Python - -```python -class Solution: - def solveNQueens(self, n: int) -> List[List[str]]: - if not n: return [] - board = [['.'] * n for _ in range(n)] - res = [] - def isVaild(board,row, col): - #判断同一列是否冲突 - for i in range(len(board)): - if board[i][col] == 'Q': - return False - # 判断左上角是否冲突 - i = row -1 - j = col -1 - while i>=0 and j>=0: - if board[i][j] == 'Q': - return False - i -= 1 - j -= 1 - # 判断右上角是否冲突 - i = row - 1 - j = col + 1 - while i>=0 and j < len(board): - if board[i][j] == 'Q': - return False - i -= 1 - j += 1 - return True - - def backtracking(board, row, n): - # 如果走到最后一行，说明已经找到一个解 - if row == n: - temp_res = [] - for temp in board: - temp_str = "".join(temp) - temp_res.append(temp_str) - res.append(temp_res) - for col in range(n): - if not isVaild(board, row, col): - continue - board[row][col] = 'Q' - backtracking(board, row+1, n) - board[row][col] = '.' - backtracking(board, 0, n) - return res -``` - ### Java ```java @@ -343,6 +293,55 @@ class Solution { } ``` +### Python + +```python +class Solution: + def solveNQueens(self, n: int) -> List[List[str]]: + if not n: return [] + board = [['.'] * n for _ in range(n)] + res = [] + def isVaild(board,row, col): + #判断同一列是否冲突 + for i in range(len(board)): + if board[i][col] == 'Q': + return False + # 判断左上角是否冲突 + i = row -1 + j = col -1 + while i>=0 and j>=0: + if board[i][j] == 'Q': + return False + i -= 1 + j -= 1 + # 判断右上角是否冲突 + i = row - 1 + j = col + 1 + while i>=0 and j < len(board): + if board[i][j] == 'Q': + return False + i -= 1 + j += 1 + return True + + def backtracking(board, row, n): + # 如果走到最后一行，说明已经找到一个解 + if row == n: + temp_res = [] + for temp in board: + temp_str = "".join(temp) + temp_res.append(temp_str) + res.append(temp_res) + for col in range(n): + if not isVaild(board, row, col): + continue + board[row][col] = 'Q' + backtracking(board, row+1, n) + board[row][col] = '.' + backtracking(board, 0, n) + return res +``` + ### Go ```Go @@ -398,6 +397,8 @@ func isValid(n, row, col int, chessboard [][]string) bool { return true } ``` + + ### Javascript ```Javascript var solveNQueens = function(n) { From 300b8f9f3e5e0233947cb08d97ea67f4840fceb3 Mon Sep 17 00:00:00 2001 From: Yuki Chen <826992207@qq.com> Date: Thu, 18 Aug 2022 10:52:42 +0800 Subject: [PATCH 03/32] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200035.=E6=90=9C?= =?UTF-8?q?=E7=B4=A2=E6=8F=92=E5=85=A5=E4=BD=8D=E7=BD=AE=20=E7=AC=AC?= =?UTF-8?q?=E4=BA=8C=E7=A7=8D=E4=BA=8C=E5=88=86=E6=B3=95Java=E7=89=88?= =?UTF-8?q?=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0035.搜索插入位置.md | 25 +++++++++++++++++++++++++ 1 file changed, 25 insertions(+) diff --git a/problems/0035.搜索插入位置.md b/problems/0035.搜索插入位置.md index f2bbb8ca..5ed3ac56 100644 --- a/problems/0035.搜索插入位置.md +++ b/problems/0035.搜索插入位置.md @@ -226,7 +226,32 @@ class Solution { } } ``` +```java +//第二种二分法：左闭右开 +public int searchInsert(int[] nums, int target) { + int left = 0; + int right = nums.length; + while (left < right) { //左闭右开 [left, right) + int middle = left + ((right - left) >> 1); + if (nums[middle] > target) { + right = middle; // target 在左区间，在[left, middle)中 + } else if (nums[middle] < target) { + left = middle + 1; // target 在右区间，在 [middle+1, right)中 + } else { // nums[middle] == target + return middle; // 数组中找到目标值的情况，直接返回下标 + } + } + // 目标值在数组所有元素之前 [0,0) + // 目标值插入数组中的位置 [left, right) ，return right 即可 + // 目标值在数组所有元素之后的情况 [left, right)，因为是右开区间，所以 return right + return right; +} +``` + + + Golang: + ```golang // 第一种二分法 func searchInsert(nums []int, target int) int { From 708c3eefdf506d2e99df63e44826877980c981cf Mon Sep 17 00:00:00 2001 From: "xin.liu" Date: Thu, 18 Aug 2022 11:08:50 +0800 Subject: [PATCH 04/32] =?UTF-8?q?Update=200704.=E4=BA=8C=E5=88=86=E6=B3=95?= =?UTF-8?q?=E6=9F=A5=E6=89=BE=20JavaScript=E5=92=8CTypeScript=E7=89=88?= =?UTF-8?q?=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0704.二分查找.md | 20 ++++++++++++-------- 1 file changed, 12 insertions(+), 8 deletions(-) diff --git a/problems/0704.二分查找.md b/problems/0704.二分查找.md index eec97a78..53f5331a 100644 --- a/problems/0704.二分查找.md +++ b/problems/0704.二分查找.md @@ -290,10 +290,11 @@ func search(nums []int, target int) int { */ var search = function(nums, target) { // right是数组最后一个数的下标，num[right]在查找范围内，是左闭右闭区间 - let left = 0, right = nums.length - 1; + let mid, left = 0, right = nums.length - 1; // 当left=right时，由于nums[right]在查找范围内，所以要包括此情况 while (left <= right) { - let mid = left + Math.floor((right - left)/2); + // 位运算 + 防止大数溢出 + mid = left + ((right - left) >> 1); // 如果中间数大于目标值，要把中间数排除查找范围，所以右边界更新为mid-1；如果右边界更新为mid，那中间数还在下次查找范围内 if (nums[mid] > target) { right = mid - 1; // 去左面闭区间寻找 @@ -316,10 +317,11 @@ var search = function(nums, target) { */ var search = function(nums, target) { // right是数组最后一个数的下标+1，nums[right]不在查找范围内，是左闭右开区间 - let left = 0, right = nums.length; + let mid, left = 0, right = nums.length; // 当left=right时，由于nums[right]不在查找范围，所以不必包括此情况 while (left < right) { - let mid = left + Math.floor((right - left)/2); + // 位运算 + 防止大数溢出 + mid = left + ((right - left) >> 1); // 如果中间值大于目标值，中间值不应在下次查找的范围内，但中间值的前一个值应在； // 由于right本来就不在查找范围内，所以将右边界更新为中间值，如果更新右边界为mid-1则将中间值的前一个值也踢出了下次寻找范围 if (nums[mid] > target) { @@ -340,9 +342,10 @@ var search = function(nums, target) { ```typescript function search(nums: number[], target: number): number { - let left: number = 0, right: number = nums.length - 1; + let mid: number, left: number = 0, right: number = nums.length - 1; while (left <= right) { - let mid: number = left + Math.floor((right - left) / 2); + // 位运算 + 防止大数溢出 + mid = left + ((right - left) >> 1); if (nums[mid] > target) { right = mid - 1; } else if (nums[mid] < target) { @@ -359,9 +362,10 @@ function search(nums: number[], target: number): number { ```typescript function search(nums: number[], target: number): number { - let left: number = 0, right: number = nums.length; + let mid: number, left: number = 0, right: number = nums.length; while (left < right) { - let mid: number = left + Math.floor((right - left) / 2); + // 位运算 + 防止大数溢出 + mid = left +((right - left) >> 1); if (nums[mid] > target) { right = mid; } else if (nums[mid] < target) { From 3fdf4484ef93ba60e5a7de589972c2c0edf2dc1c Mon Sep 17 00:00:00 2001 From: Yukun J Date: Wed, 17 Aug 2022 22:43:28 -0700 Subject: [PATCH 05/32] =?UTF-8?q?Update:=20=E5=A2=9E=E5=8A=A0<131.?= =?UTF-8?q?=E5=88=86=E5=89=B2=E5=9B=9E=E6=96=87=E4=B8=B2>=E4=BC=98?= =?UTF-8?q?=E5=8C=96=E4=BB=A3=E7=A0=81?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0131.分割回文串.md | 59 ++++++++++++++++++++++++++++++++ 1 file changed, 59 insertions(+) diff --git a/problems/0131.分割回文串.md b/problems/0131.分割回文串.md index a54d6576..37132503 100644 --- a/problems/0131.分割回文串.md +++ b/problems/0131.分割回文串.md @@ -206,6 +206,65 @@ public: return result; } }; +``` +# 优化 + +上面的代码还存在一定的优化空间, 在于如何更高效的计算一个子字符串是否是回文字串。上述代码```isPalindrome```函数运用双指针的方法来判定对于一个字符串```s```, 给定起始下标和终止下标, 截取出的子字符串是否是回文字串。但是其中有一定的重复计算存在: + +例如给定字符串```"abcde"```, 在已知```"bcd"```不是回文字串时, 不再需要去双指针操作```"abcde"```而可以直接判定它一定不是回文字串。 + +具体来说, 给定一个字符串`s`, 长度为```n```, 它成为回文字串的充分必要条件是```s[0] == s[n-1]```且```s[1:n-1]```是回文字串。 + +大家如果熟悉动态规划这种算法的话, 我们可以高效地事先一次性计算出, 针对一个字符串```s```, 它的任何子串是否是回文字串, 然后在我们的回溯函数中直接查询即可, 省去了双指针移动判定这一步骤. + +具体参考代码如下: + +```CPP +class Solution { +private: + vector> result; + vector path; // 放已经回文的子串 + vector> isPalindrome; // 放事先计算好的是否回文子串的结果 + void backtracking (const string& s, int startIndex) { + // 如果起始位置已经大于s的大小，说明已经找到了一组分割方案了 + if (startIndex >= s.size()) { + result.push_back(path); + return; + } + for (int i = startIndex; i < s.size(); i++) { + if (isPalindrome[startIndex][i]) { // 是回文子串 + // 获取[startIndex,i]在s中的子串 + string str = s.substr(startIndex, i - startIndex + 1); + path.push_back(str); + } else { // 不是回文，跳过 + continue; + } + backtracking(s, i + 1); // 寻找i+1为起始位置的子串 + path.pop_back(); // 回溯过程，弹出本次已经填在的子串 + } + } + void computePalindrome(const string& s) { + // isPalindrome[i][j] 代表 s[i:j](双边包括)是否是回文字串 + isPalindrome.resize(s.size(), vector(s.size(), false)); // 根据字符串s, 刷新布尔矩阵的大小 + for (int i = s.size() - 1; i >= 0; i--) { + // 需要倒序计算, 保证在i行时, i+1行已经计算好了 + for (int j = i; j < s.size(); j++) { + if (j == i) {isPalindrome[i][j] = true;} + else if (j - i == 1) {isPalindrome[i][j] = (s[i] == s[j]);} + else {isPalindrome[i][j] = (s[i] == s[j] && isPalindrome[i+1][j-1]);} + } + } + } +public: + vector> partition(string s) { + result.clear(); + path.clear(); + computePalindrome(s); + backtracking(s, 0); + return result; + } +}; + ``` # 总结 From 642734522dcb0aa693ab651a6591fd315a721db2 Mon Sep 17 00:00:00 2001 From: ekertree Date: Thu, 18 Aug 2022 20:07:34 +0800 Subject: [PATCH 06/32] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200844=E6=AF=94?= =?UTF-8?q?=E8=BE=83=E5=90=AB=E9=80=80=E6=A0=BC=E7=9A=84=E5=AD=97=E7=AC=A6?= =?UTF-8?q?=E4=B8=B2=20Java=20=E7=89=88=E6=9C=AC=20=E5=88=A0=E9=99=A4=20Ja?= =?UTF-8?q?va=E6=A0=87=E9=A2=98=E7=9A=84=E5=86=92=E5=8F=B7=20=E7=BB=9F?= =?UTF-8?q?=E4=B8=80=E9=A1=B5=E9=9D=A2=E6=A0=BC=E5=BC=8F?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0844.比较含退格的字符串.md | 75 +++++++++++++++++++- 1 file changed, 74 insertions(+), 1 deletion(-) diff --git a/problems/0844.比较含退格的字符串.md b/problems/0844.比较含退格的字符串.md index f7823e59..5d29ffb0 100644 --- a/problems/0844.比较含退格的字符串.md +++ b/problems/0844.比较含退格的字符串.md @@ -157,7 +157,7 @@ public: ## 其他语言版本 -### Java： +### Java ```java // 普通方法（使用栈的思路） @@ -214,7 +214,80 @@ public static boolean backspaceCompare(String s, String t) { } ``` +```java +class Solution { + public static boolean backspaceCompare(String s, String t) { + return getStr(s).equals(getStr(t)); + } + public static String getStr(String s) { //使用快慢双指针去除字符串中的# + int slowIndex; + int fastIndex = 0; + StringBuilder builder = new StringBuilder(s); //StringBuilder用于修改字符串 + for(slowIndex = 0; fastIndex < s.length(); fastIndex++) { + if(builder.charAt(fastIndex) != '#') { + builder.setCharAt(slowIndex++,builder.charAt(fastIndex)); + } else { + if(slowIndex > 0) { + slowIndex--; + } + } + } + return builder.toString().substring(0,slowIndex); //截取有效字符串 + } +} +``` + +从后往前双指针： + +```java +class Solution { + public static boolean backspaceCompare(String s, String t) { + int sSkipNum = 0; //记录s的#的个数 + int tSkipNum = 0; //记录t的#的个数 + int sIndex = s.length() - 1; + int tIndex = t.length() - 1; + while(true) { + while(sIndex >= 0) { //每次记录连续的#并跳过被删除的字符 + if(s.charAt(sIndex) == '#') { + sSkipNum++; + } else { + if(sSkipNum > 0) { + sSkipNum--; + } else { + break; + } + } + sIndex--; + } + while(tIndex >= 0) { //每次记录连续的#并跳过被删除的字符 + if(t.charAt(tIndex) == '#') { + tSkipNum++; + } else { + if(tSkipNum > 0) { + tSkipNum--; + } else { + break; + } + } + tIndex--; + } + if(sIndex < 0 || tIndex < 0) { //s 或者 t遍历完了 + break; + } + if(s.charAt(sIndex) != t.charAt(tIndex)) { //当前下标的字符不相等 + return false; + } + sIndex--; + tIndex--; + } + if(sIndex == -1 && tIndex == -1) { //同时遍历完 则相等 + return true; + } + return false; + } +} +``` ### python From 058f64624a062083aeb78ea5589c6e22ce444a12 Mon Sep 17 00:00:00 2001 From: wang2jun <91008685+wang2jun@users.noreply.github.com> Date: Fri, 19 Aug 2022 10:42:17 +0800 Subject: [PATCH 07/32] =?UTF-8?q?0100=E7=9B=B8=E5=90=8C=E7=9A=84=E6=A0=91?= =?UTF-8?q?=20=E5=A2=9E=E5=8A=A0=20js=20=E8=BF=AD=E4=BB=A3=E6=B3=95?= =?UTF-8?q?=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0100.相同的树.md | 22 ++++++++++++++++++++++ 1 file changed, 22 insertions(+) diff --git a/problems/0100.相同的树.md b/problems/0100.相同的树.md index 4b6eb7aa..9820173d 100644 --- a/problems/0100.相同的树.md +++ b/problems/0100.相同的树.md @@ -253,6 +253,28 @@ var isSameTree = function (p, q) { return isSameTree(p.left, q.left) && isSameTree(p.right, q.right); }; ``` +> 迭代法 + +```javascript +var isSameTree = (p, q) => { + const queue = [{ p, q }]; + // 这是用{ } 解决了null的问题！ + while (queue.length) { + const cur = queue.shift(); + if (cur.p == null && cur.q == null) continue; + if (cur.p == null || cur.q == null) return false; + if (cur.p.val != cur.q.val) return false; + queue.push({ + p: cur.p.left, + q: cur.q.left + }, { + p: cur.p.right, + q: cur.q.right + }); + } + return true; +}; +``` TypeScript: From 10734b6f20a881532b74f6313023717830e756dd Mon Sep 17 00:00:00 2001 From: marspere <1587393449@qq.com> Date: Fri, 19 Aug 2022 10:50:33 +0800 Subject: [PATCH 08/32] =?UTF-8?q?=E4=BC=98=E5=8C=96=E4=BA=8C=E5=8F=89?= =?UTF-8?q?=E6=A0=91=E5=B1=82=E5=BA=8F=E9=81=8D=E5=8E=86116=E5=92=8C117?= =?UTF-8?q?=E9=A2=98=E7=9A=84go=E6=96=B9=E6=B3=95?= 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 | 69 +++++++++++------------ 1 file changed, 33 insertions(+), 36 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 9ad34494..69f2f594 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 @@ -1891,36 +1891,35 @@ go: */ func connect(root *Node) *Node { - res:=[][]*Node{} - if root==nil{//防止为空 + if root == nil { //防止为空 return root } - queue:=list.New() + queue := list.New() queue.PushBack(root) - var tmpArr []*Node - for queue.Len()>0 { - length:=queue.Len()//保存当前层的长度，然后处理当前层（十分重要，防止添加下层元素影响判断层中元素的个数） - for i:=0;i 0 { + length := queue.Len() //保存当前层的长度，然后处理当前层（十分重要，防止添加下层元素影响判断层中元素的个数） + for i := 0; i < length; i++ { + node := queue.Remove(queue.Front()).(*Node) //出队列 + if node.Left != nil { queue.PushBack(node.Left) } - if node.Right!=nil{ + if node.Right != nil { queue.PushBack(node.Right) } - tmpArr=append(tmpArr,node)//将值加入本层切片中 + tmpArr = append(tmpArr, node) //将值加入本层切片中 } - res=append(res,tmpArr)//放入结果集 - tmpArr=[]*Node{}//清空层的数据 - } - //遍历每层元素,指定next - for i:=0;i 1 { + // 遍历每层元素,指定next + for i := 0; i < len(tmpArr)-1; i++ { + tmpArr[i].Next = tmpArr[i+1] + } } + tmpArr = []*Node{} //清空层的数据 } return root } + ``` Swift： @@ -2172,33 +2171,31 @@ go: */ func connect(root *Node) *Node { - res:=[][]*Node{} - if root==nil{//防止为空 + if root == nil { //防止为空 return root } - queue:=list.New() + queue := list.New() queue.PushBack(root) - var tmpArr []*Node - for queue.Len()>0 { - length:=queue.Len()//保存当前层的长度，然后处理当前层（十分重要，防止添加下层元素影响判断层中元素的个数） - for i:=0;i 0 { + length := queue.Len() //保存当前层的长度，然后处理当前层（十分重要，防止添加下层元素影响判断层中元素的个数） + for i := 0; i < length; i++ { + node := queue.Remove(queue.Front()).(*Node) //出队列 + if node.Left != nil { queue.PushBack(node.Left) } - if node.Right!=nil{ + if node.Right != nil { queue.PushBack(node.Right) } - tmpArr=append(tmpArr,node)//将值加入本层切片中 + tmpArr = append(tmpArr, node) //将值加入本层切片中 } - res=append(res,tmpArr)//放入结果集 - tmpArr=[]*Node{}//清空层的数据 - } - //遍历每层元素,指定next - for i:=0;i 1 { + // 遍历每层元素,指定next + for i := 0; i < len(tmpArr)-1; i++ { + tmpArr[i].Next = tmpArr[i+1] + } } + tmpArr = []*Node{} //清空层的数据 } return root } From 4f71ddc10fdacabdb14acfd53cbcee7bb0d7e996 Mon Sep 17 00:00:00 2001 From: marspere <1587393449@qq.com> Date: Fri, 19 Aug 2022 11:28:57 +0800 Subject: [PATCH 09/32] =?UTF-8?q?Update=200102:=E4=BC=98=E5=8C=96=E4=BA=8C?= =?UTF-8?q?=E5=8F=89=E6=A0=91=E5=B1=82=E5=BA=8F=E9=81=8D=E5=8E=86=E4=B8=AD?= =?UTF-8?q?515=E9=A2=98=E7=9A=84go=E6=96=B9=E6=B3=95?= 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 | 52 +++++++++++------------ 1 file changed, 24 insertions(+), 28 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 69f2f594..a1a7baed 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 @@ -1552,44 +1552,40 @@ go: 515. 在每个树行中找最大值 */ func largestValues(root *TreeNode) []int { - res:=[][]int{} - var finRes []int - if root==nil{//防止为空 - return finRes + if root == nil { + //防止为空 + return nil } - queue:=list.New() + queue := list.New() queue.PushBack(root) - var tmpArr []int - //层次遍历 - for queue.Len()>0 { - length:=queue.Len()//保存当前层的长度，然后处理当前层（十分重要，防止添加下层元素影响判断层中元素的个数） - for i:=0;i 0 { + //保存当前层的长度，然后处理当前层（十分重要，防止添加下层元素影响判断层中元素的个数） + length := queue.Len() + for i := 0; i < length; i++ { + node := queue.Remove(queue.Front()).(*TreeNode)//出队列 + // 比较当前层中的最大值和新遍历的元素大小，取两者中大值 + temp = max(temp, node.Val) + if node.Left != nil { queue.PushBack(node.Left) } - if node.Right!=nil{ + if node.Right != nil { queue.PushBack(node.Right) } - tmpArr=append(tmpArr,node.Val)//将值加入本层切片中 } - res=append(res,tmpArr)//放入结果集 - tmpArr=[]int{}//清空层的数据 + ans = append(ans, temp) + temp = math.MinInt64 } - //找到每层的最大值 - for i:=0;i max { - max = val - } + +func max(x, y int) int { + if x > y { + return x } - return max + return y } ``` From 33c079738631c552185ba8476c893b90ecf477d0 Mon Sep 17 00:00:00 2001 From: marspere <1587393449@qq.com> Date: Fri, 19 Aug 2022 13:50:48 +0800 Subject: [PATCH 10/32] =?UTF-8?q?Update=200102:=E4=BC=98=E5=8C=96=E4=BA=8C?= =?UTF-8?q?=E5=8F=89=E6=A0=91=E7=9A=84=E5=B1=82=E5=BA=8F=E9=81=8D=E5=8E=86?= =?UTF-8?q?=E4=B8=AD637=E9=A2=98=E7=9A=84go=E6=96=B9=E6=B3=95?= 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, 20 insertions(+), 26 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 a1a7baed..25189dba 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 @@ -1054,40 +1054,34 @@ go: 637. 二叉树的层平均值 */ func averageOfLevels(root *TreeNode) []float64 { - res:=[][]int{} - var finRes []float64 - if root==nil{//防止为空 - return finRes + if root == nil { + // 防止为空 + return nil } - queue:=list.New() + res := make([]float64, 0) + queue := list.New() queue.PushBack(root) - var tmpArr []int - for queue.Len()>0 { - length:=queue.Len()//保存当前层的长度，然后处理当前层（十分重要，防止添加下层元素影响判断层中元素的个数） - for i:=0;i 0 { + //保存当前层的长度，然后处理当前层（十分重要，防止添加下层元素影响判断层中元素的个数） + length := queue.Len() + for i := 0; i < length; i++ { + node := queue.Remove(queue.Front()).(*TreeNode) + if node.Left != nil { queue.PushBack(node.Left) } - if node.Right!=nil{ + if node.Right != nil { queue.PushBack(node.Right) } - tmpArr=append(tmpArr,node.Val)//将值加入本层切片中 + // 当前层元素求和 + sum += node.Val } - res=append(res,tmpArr)//放入结果集 - tmpArr=[]int{}//清空层的数据 + // 计算每层的平均值，将结果添加到响应结果中 + res = append(res, float64(sum)/float64(length)) + sum = 0 // 清空该层的数据 } - //计算每层的平均值 - length:=len(res) - for i:=0;i Date: Fri, 19 Aug 2022 13:56:47 +0800 Subject: [PATCH 11/32] =?UTF-8?q?Update=200102:=E4=BC=98=E5=8C=96=E4=BA=8C?= =?UTF-8?q?=E5=8F=89=E6=A0=91=E7=9A=84=E5=B1=82=E5=BA=8F=E9=81=8D=E5=8E=86?= =?UTF-8?q?=E4=B8=AD199=E9=A2=98=E7=9A=84go=E6=96=B9=E6=B3=95?= 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 | 35 +++++++++++------------ 1 file changed, 16 insertions(+), 19 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 25189dba..0063f36e 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 @@ -810,33 +810,30 @@ go: 199. 二叉树的右视图 */ func rightSideView(root *TreeNode) []int { - queue:=list.New() - res:=[][]int{} - var finaRes []int - if root==nil{ - return finaRes + if root == nil { + return nil } + res := make([]int, 0) + queue := list.New() queue.PushBack(root) - for queue.Len()>0{ - length:=queue.Len() - tmp:=[]int{} - for i:=0;i 0 { + length := queue.Len() + for i := 0; i < length; i++ { + node := queue.Remove(queue.Front()).(*TreeNode) + if node.Left != nil { queue.PushBack(node.Left) } - if node.Right!=nil{ + if node.Right != nil { queue.PushBack(node.Right) } - tmp=append(tmp,node.Val) + // 取每层的最后一个元素，添加到结果集中 + if i == length-1 { + res = append(res, node.Val) + } } - res=append(res,tmp) } - //取每一层的最后一个元素 - for i:=0;i Date: Fri, 19 Aug 2022 16:52:06 +0800 Subject: [PATCH 12/32] =?UTF-8?q?=E6=9B=B4=E6=96=B00129.=E6=B1=82=E6=A0=B9?= =?UTF-8?q?=E5=88=B0=E5=8F=B6=E5=AD=90=E7=BB=93=E7=82=B9=E6=95=B0=E5=AD=97?= =?UTF-8?q?=E4=B9=8B=E5=92=8C=E4=B8=AD=EF=BC=8Cgo=E8=AF=AD=E8=A8=80?= =?UTF-8?q?=E4=BB=A3=E7=A0=81=E6=97=A0=E6=B3=95=E9=80=9A=E8=BF=87leetcode?= =?UTF-8?q?=E7=9A=84=E8=AF=AD=E6=B3=95=E9=94=99=E8=AF=AF?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 修改了原本的go代码的错误（原本的go代码无法通过leetcode的题目执行，有语法错误） --- .../0129.求根到叶子节点数字之和.md | 28 +++++++++---------- 1 file changed, 14 insertions(+), 14 deletions(-) diff --git a/problems/0129.求根到叶子节点数字之和.md b/problems/0129.求根到叶子节点数字之和.md index 92a72fe3..859df39e 100644 --- a/problems/0129.求根到叶子节点数字之和.md +++ b/problems/0129.求根到叶子节点数字之和.md @@ -250,22 +250,22 @@ Go： ```go func sumNumbers(root *TreeNode) int { - sum = 0 - travel(root, root.Val) - return sum + sum := 0 + dfs(root, root.Val, &sum) + return sum } -func travel(root *TreeNode, tmpSum int) { - if root.Left == nil && root.Right == nil { - sum += tmpSum - } else { - if root.Left != nil { - travel(root.Left, tmpSum*10+root.Left.Val) - } - if root.Right != nil { - travel(root.Right, tmpSum*10+root.Right.Val) - } - } +func dfs(root *TreeNode, tmpSum int, sum *int) { + if root.Left == nil && root.Right == nil { + *sum += tmpSum + } else { + if root.Left != nil { + dfs(root.Left, tmpSum*10 + root.Left.Val, sum) + } + if root.Right != nil { + dfs(root.Right, tmpSum*10 + root.Right.Val, sum) + } + } } ``` From 87a51bcb6916834fa98dc6f2d3dd1f0df69a6ac9 Mon Sep 17 00:00:00 2001 From: SwaggyP <1352164869@qq.com> Date: Sat, 20 Aug 2022 09:53:47 +0800 Subject: [PATCH 13/32] =?UTF-8?q?=E5=A2=9E=E5=8A=A00100.=E7=9B=B8=E5=90=8C?= =?UTF-8?q?=E7=9A=84=E6=A0=91go=E8=AF=AD=E8=A8=80=E7=9A=84=E9=80=92?= =?UTF-8?q?=E5=BD=92=E8=A7=A3=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 增加了0100.相同的树go语言的递归解法 --- problems/0100.相同的树.md | 20 ++++++++++++++++++++ 1 file changed, 20 insertions(+) diff --git a/problems/0100.相同的树.md b/problems/0100.相同的树.md index 4b6eb7aa..0057a85e 100644 --- a/problems/0100.相同的树.md +++ b/problems/0100.相同的树.md @@ -237,6 +237,26 @@ class Solution: return True ``` Go： +> 递归法 +```go +func isSameTree(p *TreeNode, q *TreeNode) bool { + if p != nil && q == nil { + return false + } + if p == nil && q != nil { + return false + } + if p == nil && q == nil { + return true + } + if p.Val != q.Val { + return false + } + Left := isSameTree(p.Left, q.Left) + Right := isSameTree(p.Right, q.Right) + return Left && Right +} +``` JavaScript： From 6a201720030faa7a70abcc313c1aa756245f3329 Mon Sep 17 00:00:00 2001 From: AronJudge <2286381138@qq.com> Date: Sat, 20 Aug 2022 10:38:58 +0800 Subject: [PATCH 14/32] =?UTF-8?q?463=20=E5=B2=9B=E5=B1=BF=E7=9A=84?= =?UTF-8?q?=E5=91=A8=E9=95=BF?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 更新Java的方法二 --- problems/0463.岛屿的周长.md | 21 +++++++++++++++++++++ 1 file changed, 21 insertions(+) diff --git a/problems/0463.岛屿的周长.md b/problems/0463.岛屿的周长.md index ea038140..3dc69f20 100644 --- a/problems/0463.岛屿的周长.md +++ b/problems/0463.岛屿的周长.md @@ -118,6 +118,27 @@ class Solution { return res; } } + +// 解法二 +class Solution { + public int islandPerimeter(int[][] grid) { + // 计算岛屿的周长 + // 方法二 : 遇到相邻的陆地总周长就-2 + int landSum = 0; // 陆地数量 + int cover = 0; // 相邻陆地数量 + for (int i = 0; i < grid.length; i++) { + for (int j = 0; j < grid[0].length; j++) { + if (grid[i][j] == 1) { + landSum++; + // 统计上面和左边的相邻陆地 + if(i - 1 >= 0 && grid[i-1][j] == 1) cover++; + if(j - 1 >= 0 && grid[i][j-1] == 1) cover++; + } + } + } + return landSum * 4 - cover * 2; + } +} ``` Python： From d53246138d67e2054b54bd30b596cc4e427ee25a Mon Sep 17 00:00:00 2001 From: Yuki Chen <826992207@qq.com> Date: Sat, 20 Aug 2022 11:54:39 +0800 Subject: [PATCH 15/32] =?UTF-8?q?=E6=B7=BB=E5=8A=A00027.=E7=A7=BB=E9=99=A4?= =?UTF-8?q?=E5=85=83=E7=B4=A0=20=E7=9B=B8=E5=90=91=E5=8F=8C=E6=8C=87?= =?UTF-8?q?=E9=92=88=E6=B3=95Java=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0027.移除元素.md | 26 +++++++++++++++++++++----- 1 file changed, 21 insertions(+), 5 deletions(-) diff --git a/problems/0027.移除元素.md b/problems/0027.移除元素.md index 12f88c75..68b1d4bd 100644 --- a/problems/0027.移除元素.md +++ b/problems/0027.移除元素.md @@ -146,7 +146,7 @@ public: }; ``` - + ## 相关题目推荐 * 26.删除排序数组中的重复项 @@ -154,10 +154,6 @@ public: * 844.比较含退格的字符串 * 977.有序数组的平方 - - - - ## 其他语言版本 @@ -180,6 +176,26 @@ class Solution { } } ``` +```java +//相向双指针法 +class Solution { + public int removeElement(int[] nums, int val) { + int left = 0; + int right = nums.length - 1; + while(right >= 0 && nums[right] == val) right--; //将right移到从右数第一个值不为val的位置 + while(left <= right) { + if(nums[left] == val) { //left位置的元素需要移除 + //将right位置的元素移到left（覆盖），right位置移除 + nums[left] = nums[right]; + right--; + } + left++; + while(right >= 0 && nums[right] == val) right--; + } + return left; + } +} +``` Python： From b2534f7993018e497670e05fdf258830a1f4b69b Mon Sep 17 00:00:00 2001 From: AronJudge <2286381138@qq.com> Date: Sat, 20 Aug 2022 16:50:29 +0800 Subject: [PATCH 16/32] =?UTF-8?q?05=20=E6=9C=80=E9=95=BF=E5=9B=9E=E6=96=87?= =?UTF-8?q?=E5=AD=90=E4=B8=B2?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 更新Java动态绘画方法， 之前的根本就不是这个题 --- problems/0005.最长回文子串.md | 44 ++++++++++++++++++++--------- 1 file changed, 31 insertions(+), 13 deletions(-) diff --git a/problems/0005.最长回文子串.md b/problems/0005.最长回文子串.md index d53acf63..893525d2 100644 --- a/problems/0005.最长回文子串.md +++ b/problems/0005.最长回文子串.md @@ -263,21 +263,39 @@ public: Java： ```java -public int[] twoSum(int[] nums, int target) { - int[] res = new int[2]; - if(nums == null || nums.length == 0){ - return res; - } - Map map = new HashMap<>(); - for(int i = 0; i < nums.length; i++){ - int temp = target - nums[i]; - if(map.containsKey(temp)){ - res[1] = i; - res[0] = map.get(temp); +// 双指针 动态规划 +class Solution { + public String longestPalindrome(String s) { + if (s.length() == 0 || s.length() == 1) return s; + int length = 1; + int index = 0; + boolean[][] palindrome = new boolean[s.length()][s.length()]; + for (int i = 0; i < s.length(); i++) { + palindrome[i][i] = true; } - map.put(nums[i], i); + + for (int L = 2; L <= s.length(); L++) { + for (int i = 0; i < s.length(); i++) { + int j = i + L - 1; + if (j >= s.length()) break; + + if (s.charAt(i) != s.charAt(j)) { + palindrome[i][j] = false; + } else { + if (j - i < 3) { + palindrome[i][j] = true; + } else { + palindrome[i][j] = palindrome[i + 1][j - 1]; + } + } + if (palindrome[i][j] && j - i + 1 > length) { + length = j - i + 1; + index = i; + } + } + } + return s.substring(index, index + length); } - return res; } ``` From a97a7ef2ece2b73f87c196e01037508d1f197deb Mon Sep 17 00:00:00 2001 From: AronJudge <2286381138@qq.com> Date: Sat, 20 Aug 2022 16:50:29 +0800 Subject: [PATCH 17/32] =?UTF-8?q?05=20=E6=9C=80=E9=95=BF=E5=9B=9E=E6=96=87?= =?UTF-8?q?=E5=AD=90=E4=B8=B2?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 更新Java动态规划方法 之前的根本就不是这个题 --- problems/0005.最长回文子串.md | 43 ++++++++++++++++++++--------- 1 file changed, 30 insertions(+), 13 deletions(-) diff --git a/problems/0005.最长回文子串.md b/problems/0005.最长回文子串.md index d53acf63..b9b568ab 100644 --- a/problems/0005.最长回文子串.md +++ b/problems/0005.最长回文子串.md @@ -263,21 +263,38 @@ public: Java： ```java -public int[] twoSum(int[] nums, int target) { - int[] res = new int[2]; - if(nums == null || nums.length == 0){ - return res; - } - Map map = new HashMap<>(); - for(int i = 0; i < nums.length; i++){ - int temp = target - nums[i]; - if(map.containsKey(temp)){ - res[1] = i; - res[0] = map.get(temp); +// 双指针 动态规划 +class Solution { + public String longestPalindrome(String s) { + if (s.length() == 0 || s.length() == 1) return s; + int length = 1; + int index = 0; + boolean[][] palindrome = new boolean[s.length()][s.length()]; + for (int i = 0; i < s.length(); i++) { + palindrome[i][i] = true; } - map.put(nums[i], i); + + for (int L = 2; L <= s.length(); L++) { + for (int i = 0; i < s.length(); i++) { + int j = i + L - 1; + if (j >= s.length()) break; + if (s.charAt(i) != s.charAt(j)) { + palindrome[i][j] = false; + } else { + if (j - i < 3) { + palindrome[i][j] = true; + } else { + palindrome[i][j] = palindrome[i + 1][j - 1]; + } + } + if (palindrome[i][j] && j - i + 1 > length) { + length = j - i + 1; + index = i; + } + } + } + return s.substring(index, index + length); } - return res; } ``` From 8e3524d482764d29e3237cc85bc6daceb4c82b53 Mon Sep 17 00:00:00 2001 From: ekertree Date: Sat, 20 Aug 2022 21:02:18 +0800 Subject: [PATCH 18/32] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=2054.=E8=9E=BA?= =?UTF-8?q?=E6=97=8B=E7=9F=A9=E9=98=B5=20Java=20=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0054.螺旋矩阵.md | 68 +++++++++++++++++++++++++++++++++++ 1 file changed, 68 insertions(+) diff --git a/problems/0054.螺旋矩阵.md b/problems/0054.螺旋矩阵.md index efbda5ff..aeb8e521 100644 --- a/problems/0054.螺旋矩阵.md +++ b/problems/0054.螺旋矩阵.md @@ -133,5 +133,73 @@ public: ## 其他语言版本 +### Java + +```java +class Solution { + public List spiralOrder(int[][] matrix) { + //存放数组的数 + List ans = new ArrayList<>(); + //列数 + int columns = matrix[0].length; + //行数 + int rows = matrix.length; + //遍历起点 + int start = 0; + //循环的次数 行数和列数中的最小值除以二 + int loop = Math.min(rows,columns) / 2; + //未遍历的中间列（行）的列（行）下标 + int mid = loop; + //终止条件 + int offSet = 1; + int i,j; + while(loop-- > 0) { + //初始化起点 + i = j = start; + + //从左往右 + for(; j < columns - offSet; j++) + ans.add(matrix[i][j]); + + //从上往下 + for(; i < rows - offSet; i++) + ans.add(matrix[i][j]); + + //从右往左 + for(; j > start; j--) + ans.add(matrix[i][j]); + + //从下往上 + for(; i > start; i--) + ans.add(matrix[i][j]); + + //每循环一次 改变起点位置 + start++; + //终止条件改变 + offSet++; + } + + //如果行和列中的最小值是奇数 则会产生中间行或者中间列没有遍历 + if(Math.min(rows,columns) % 2 != 0) { + //行大于列则产生中间列 + if(rows > columns) { + //中间列的行的最大下标的下一位的下标为mid + rows - columns + 1 + for(int k = mid; k < mid + rows - columns + 1; k++) { + ans.add(matrix[k][mid]); + } + }else {//列大于等于行则产生中间行 + //中间行的列的最大下标的下一位的下标为mid + columns - rows + 1 + for(int k = mid; k < mid + columns - rows + 1; k++) { + ans.add(matrix[mid][k]); + } + } + } + return ans; + } +} +``` + + + -----------------------
From 0c2f5c912cd628157e2e049e80abdd3c9984d8e5 Mon Sep 17 00:00:00 2001 From: Luo <82520819+Jerry-306@users.noreply.github.com> Date: Sun, 21 Aug 2022 09:27:35 +0800 Subject: [PATCH 19/32] =?UTF-8?q?0070.=E7=88=AC=E6=A5=BC=E6=A2=AFJavaScrip?= =?UTF-8?q?t=E4=BB=A3=E7=A0=81=E6=9B=B4=E6=96=B0?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 原代码貌似不是本题的哦 --- problems/0070.爬楼梯完全背包版本.md | 12 +++++++----- 1 file changed, 7 insertions(+), 5 deletions(-) diff --git a/problems/0070.爬楼梯完全背包版本.md b/problems/0070.爬楼梯完全背包版本.md index ec019e57..28439e09 100644 --- a/problems/0070.爬楼梯完全背包版本.md +++ b/problems/0070.爬楼梯完全背包版本.md @@ -187,12 +187,14 @@ func climbStairs(n int) int { JavaScript: ```javascript var climbStairs = function(n) { - const dp = new Array(n+1).fill(0); - const weight = [1,2]; + const dp = new Array(n + 1).fill(0); + const m = 2; dp[0] = 1; - for(let i = 0; i <= n; i++){ //先遍历背包 - for(let j = 0; j < weight.length; j++){ // 再遍历物品 - if(i >= weight[j]) dp[i] += dp[i-weight[j]]; + for(let i = 1; i <= n; i++){ + for(let j = 1; j <= m; j++){ + if(i >= j) { + dp[i] += dp[i - j]; + } } } return dp[n]; From 7b61605475f39edb73e1ada48a61fe55e4bbea8f Mon Sep 17 00:00:00 2001 From: Jack_ZhijieFang <56966563+laerpeeK@users.noreply.github.com> Date: Mon, 22 Aug 2022 01:22:41 +0800 Subject: [PATCH 20/32] =?UTF-8?q?docs=200541.=E5=8F=8D=E8=BD=AC=E5=AD=97?= =?UTF-8?q?=E7=AC=A6=E4=B8=B2=E2=85=A1.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 修改了该题描述，更新到leetcode最新问题描述。原先的问题描述，个人觉得对 "是否包括前2k个字符" 有理解上的困难。 --- problems/0541.反转字符串II.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/0541.反转字符串II.md b/problems/0541.反转字符串II.md index 26d0b89c..80692c0b 100644 --- a/problems/0541.反转字符串II.md +++ b/problems/0541.反转字符串II.md @@ -12,7 +12,7 @@ [力扣题目链接](https://leetcode.cn/problems/reverse-string-ii/) -给定一个字符串 s 和一个整数 k，你需要对从字符串开头算起的每隔 2k 个字符的前 k 个字符进行反转。 +给定一个字符串 s 和一个整数 k，从字符串开头算起, 每计数至 2k 个字符，就反转这 2k 个字符中的前 k 个字符。 如果剩余字符少于 k 个，则将剩余字符全部反转。 From 300922d68e3305024755fffb718f5d07a05b70cd Mon Sep 17 00:00:00 2001 From: Jamcy123 <1219502823@qq.com> Date: Tue, 23 Aug 2022 11:45:24 +0800 Subject: [PATCH 21/32] =?UTF-8?q?347.=E5=89=8DK=E4=B8=AA=E9=AB=98=E9=A2=91?= =?UTF-8?q?=E5=85=83=E7=B4=A0JS=E7=89=88=E6=9C=AC=E4=BF=AE=E6=94=B9?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0347.前K个高频元素.md | 177 ++++++++++++++------------- 1 file changed, 90 insertions(+), 87 deletions(-) diff --git a/problems/0347.前K个高频元素.md b/problems/0347.前K个高频元素.md index 1d6a358b..d0a3063a 100644 --- a/problems/0347.前K个高频元素.md +++ b/problems/0347.前K个高频元素.md @@ -263,99 +263,102 @@ func topKFrequent(nums []int, k int) []int { -javaScript: +JavaScript: ```js -/** - * @param {number[]} nums - * @param {number} k - * @return {number[]} - */ -var topKFrequent = function(nums, k) { - const map = new Map(); - - for(const num of nums) { - map.set(num, (map.get(num) || 0) + 1); - } - - // 创建小顶堆 - const priorityQueue = new PriorityQueue((a, b) => a[1] - b[1]); - - // entry 是一个长度为2的数组，0位置存储key，1位置存储value - for (const entry of map.entries()) { - priorityQueue.push(entry); - if (priorityQueue.size() > k) { - priorityQueue.pop(); +// js 没有堆 需要自己构造 +class Heap { + constructor(compareFn) { + this.compareFn = compareFn; + this.queue = []; } - } - const ret = []; + // 添加 + push(item) { + // 推入元素 + this.queue.push(item); - for(let i = priorityQueue.size() - 1; i >= 0; i--) { - ret[i] = priorityQueue.pop()[0]; - } + // 上浮 + let index = this.size() - 1; // 记录推入元素下标 + let parent = Math.floor((index - 1) / 2); // 记录父节点下标 - return ret; + while (parent >= 0 && this.compare(parent, index) > 0) { // 注意compare参数顺序 + [this.queue[index], this.queue[parent]] = [this.queue[parent], this.queue[index]]; + + // 更新下标 + index = parent; + parent = Math.floor((index - 1) / 2); + } + } + + // 获取堆顶元素并移除 + pop() { + // 堆顶元素 + const out = this.queue[0]; + + // 移除堆顶元素 填入最后一个元素 + this.queue[0] = this.queue.pop(); + + // 下沉 + let index = 0; // 记录下沉元素下标 + let left = 1; // left 是左子节点下标 left + 1 则是右子节点下标 + let searchChild = this.compare(left, left + 1) > 0 ? left + 1 : left; + + while (searchChild !== undefined && this.compare(index, searchChild) > 0) { // 注意compare参数顺序 + [this.queue[index], this.queue[searchChild]] = [this.queue[searchChild], this.queue[index]]; + + // 更新下标 + index = searchChild; + left = 2 * index + 1; + searchChild = this.compare(left, left + 1) > 0 ? left + 1 : left; + } + + return out; + } + + size() { + return this.queue.length; + } + + // 使用传入的 compareFn 比较两个位置的元素 + compare(index1, index2) { + // 处理下标越界问题 + if (this.queue[index1] === undefined) return 1; + if (this.queue[index2] === undefined) return -1; + + return this.compareFn(this.queue[index1], this.queue[index2]); + } + +} + +const topKFrequent = function (nums, k) { + const map = new Map(); + + for (const num of nums) { + map.set(num, (map.get(num) || 0) + 1); + } + + // 创建小顶堆 + const heap= new Heap((a, b) => a[1] - b[1]); + + // entry 是一个长度为2的数组，0位置存储key，1位置存储value + for (const entry of map.entries()) { + heap.push(entry); + + if (heap.size() > k) { + heap.pop(); + } + } + + // return heap.queue.map(e => e[0]); + + const res = []; + + for (let i = heap.size() - 1; i >= 0; i--) { + res[i] = heap.pop()[0]; + } + + return res; }; - - -function PriorityQueue(compareFn) { - this.compareFn = compareFn; - this.queue = []; -} - -// 添加 -PriorityQueue.prototype.push = function(item) { - this.queue.push(item); - let index = this.queue.length - 1; - let parent = Math.floor((index - 1) / 2); - // 上浮 - while(parent >= 0 && this.compare(parent, index) > 0) { - // 交换 - [this.queue[index], this.queue[parent]] = [this.queue[parent], this.queue[index]]; - index = parent; - parent = Math.floor((index - 1) / 2); - } -} - -// 获取堆顶元素并移除 -PriorityQueue.prototype.pop = function() { - const ret = this.queue[0]; - - // 把最后一个节点移到堆顶 - this.queue[0] = this.queue.pop(); - - let index = 0; - // 左子节点下标，left + 1 就是右子节点下标 - let left = 1; - let selectedChild = this.compare(left, left + 1) > 0 ? left + 1 : left; - - // 下沉 - while(selectedChild !== undefined && this.compare(index, selectedChild) > 0) { - // 交换 - [this.queue[index], this.queue[selectedChild]] = [this.queue[selectedChild], this.queue[index]]; - index = selectedChild; - left = 2 * index + 1; - selectedChild = this.compare(left, left + 1) > 0 ? left + 1 : left; - } - - return ret; -} - -PriorityQueue.prototype.size = function() { - return this.queue.length; -} - -// 使用传入的 compareFn 比较两个位置的元素 -PriorityQueue.prototype.compare = function(index1, index2) { - if (this.queue[index1] === undefined) { - return 1; - } - if (this.queue[index2] === undefined) { - return -1; - } - - return this.compareFn(this.queue[index1], this.queue[index2]); -} ``` TypeScript： From 26cdabc43f91724bd41de2f03b2d157202f9f1d6 Mon Sep 17 00:00:00 2001 From: starry0819 Date: Tue, 23 Aug 2022 14:25:21 +0800 Subject: [PATCH 22/32] =?UTF-8?q?=E6=9B=B4=E6=96=B0=200707.=E8=AE=BE?= =?UTF-8?q?=E8=AE=A1=E9=93=BE=E8=A1=A8.md=20Java=E7=A4=BA=E4=BE=8B?= =?UTF-8?q?=E4=BB=A3=E7=A0=81=E9=83=A8=E5=88=86=E5=8A=9F=E8=83=BD?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 0707.设计链表.md 中 Java示例代码中，单链表的deleteAtIndex方法进行了修改。 --- problems/0707.设计链表.md | 6 +++++- 1 file changed, 5 insertions(+), 1 deletion(-) diff --git a/problems/0707.设计链表.md b/problems/0707.设计链表.md index 00886039..42912eda 100644 --- a/problems/0707.设计链表.md +++ b/problems/0707.设计链表.md @@ -353,8 +353,12 @@ class MyLinkedList { return; } size--; + if (index == 0) { + head = head.next; + return; + } ListNode pred = head; - for (int i = 0; i < index; i++) { + for (int i = 0; i < index - 1; i++) { pred = pred.next; } pred.next = pred.next.next; From 41c480371c4a9c41f9e0a06d35810529b15c8c91 Mon Sep 17 00:00:00 2001 From: SwaggyP <1352164869@qq.com> Date: Wed, 24 Aug 2022 12:41:38 +0800 Subject: [PATCH 23/32] =?UTF-8?q?0005.=E6=9C=80=E9=95=BF=E5=9B=9E=E6=96=87?= =?UTF-8?q?=E5=AD=90=E4=B8=B2=E5=A2=9E=E5=8A=A0go=E7=9A=84dp=E8=A7=A3?= =?UTF-8?q?=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 0005.最长回文子串增加go的dp解法 --- problems/0005.最长回文子串.md | 28 ++++++++++++++++++++++++++++ 1 file changed, 28 insertions(+) diff --git a/problems/0005.最长回文子串.md b/problems/0005.最长回文子串.md index d53acf63..6e407d6e 100644 --- a/problems/0005.最长回文子串.md +++ b/problems/0005.最长回文子串.md @@ -363,6 +363,34 @@ class Solution: Go： ```go +func longestPalindrome(s string) string { + maxLen := 0 + left := 0 + length := 0 + dp := make([][]bool, len(s)) + for i := 0; i < len(s); i++ { + dp[i] = make([]bool,len(s)) + } + for i := len(s)-1; i >= 0; i-- { + for j := i; j < len(s); j++ { + if s[i] == s[j]{ + if j-i <= 1{ // 情况一和情况二 + length = j-i + dp[i][j]=true + }else if dp[i+1][j-1]{ // 情况三 + length = j-i + dp[i][j] = true + } + } + } + if length > maxLen { + maxLen = length + left = i + } + } + return s[left: left+maxLen+1] +} + ``` From ed199f1f75199ab9daf487c0c6b99d467f1cef43 Mon Sep 17 00:00:00 2001 From: zhicheng lee <904688436@qq.com> Date: Wed, 24 Aug 2022 21:31:30 +0800 Subject: [PATCH 24/32] =?UTF-8?q?=E6=9B=B4=E6=96=B0=200077.=E7=BB=84?= =?UTF-8?q?=E5=90=88=E4=BC=98=E5=8C=96.md=20cpp=E4=BB=A3=E7=A0=81=E8=AF=AD?= =?UTF-8?q?=E6=B3=95=E9=AB=98=E4=BA=AE?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 网站本文cpp代码没有语法高亮，推测网站可以识别```CPP（网站其它cpp代码均用```CPP标识），但无法识别```c++ --- problems/0077.组合优化.md | 10 +++++----- 1 file changed, 5 insertions(+), 5 deletions(-) diff --git a/problems/0077.组合优化.md b/problems/0077.组合优化.md index e336fb75..b8b5f3c1 100644 --- a/problems/0077.组合优化.md +++ b/problems/0077.组合优化.md @@ -20,7 +20,7 @@ 大家先回忆一下[77. 组合]给出的回溯法的代码： -```c++ +```CPP class Solution { private: vector> result; // 存放符合条件结果的集合 @@ -52,7 +52,7 @@ public: 在遍历的过程中有如下代码： -```c++ +```CPP for (int i = startIndex; i <= n; i++) { path.push_back(i); backtracking(n, k, i + 1); @@ -76,7 +76,7 @@ for (int i = startIndex; i <= n; i++) { **如果for循环选择的起始位置之后的元素个数 已经不足 我们需要的元素个数了，那么就没有必要搜索了**。 注意代码中i，就是for循环里选择的起始位置。 -```c++ +```CPP for (int i = startIndex; i <= n; i++) { ``` @@ -98,13 +98,13 @@ for (int i = startIndex; i <= n; i++) { 所以优化之后的for循环是： -```c++ +```CPP for (int i = startIndex; i <= n - (k - path.size()) + 1; i++) // i为本次搜索的起始位置 ``` 优化后整体代码如下： -```c++ +```CPP class Solution { private: vector> result; From 0a9653df8457535038888a57e05ba873109eccce Mon Sep 17 00:00:00 2001 From: Yukun J Date: Wed, 24 Aug 2022 11:22:43 -0400 Subject: [PATCH 25/32] =?UTF-8?q?Update:=200416=20=E4=BD=BF=E7=94=A8STL?= =?UTF-8?q?=E6=B1=82=E5=92=8C?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0416.分割等和子集.md | 7 ++----- 1 file changed, 2 insertions(+), 5 deletions(-) diff --git a/problems/0416.分割等和子集.md b/problems/0416.分割等和子集.md index 03eae8ef..9c90ea27 100644 --- a/problems/0416.分割等和子集.md +++ b/problems/0416.分割等和子集.md @@ -143,15 +143,12 @@ dp[j]的数值一定是小于等于j的。 class Solution { public: bool canPartition(vector& nums) { - int sum = 0; - + // 使用标准库函数 便捷求和 + int sum = accumulate(nums.begin(), nums.end(), 0); // dp[i]中的i表示背包内总和 // 题目中说：每个数组中的元素不会超过 100，数组的大小不会超过 200 // 总和不会大于20000，背包最大只需要其中一半，所以10001大小就可以了 vector dp(10001, 0); - for (int i = 0; i < nums.size(); i++) { - sum += nums[i]; - } if (sum % 2 == 1) return false; int target = sum / 2; From 55648dcb7905550ce30081295deee4de858fef35 Mon Sep 17 00:00:00 2001 From: SwaggyP <1352164869@qq.com> Date: Sun, 28 Aug 2022 10:26:45 +0800 Subject: [PATCH 26/32] =?UTF-8?q?0127.=E5=8D=95=E8=AF=8D=E6=8E=A5=E9=BE=99?= =?UTF-8?q?=20=E5=A2=9E=E5=8A=A0go=E8=AF=AD=E8=A8=80=E7=9A=84=E8=A7=A3?= =?UTF-8?q?=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 0127.单词接龙 增加go语言的解法 --- problems/0127.单词接龙.md | 51 +++++++++++++++++++++++++++++++++++ 1 file changed, 51 insertions(+) diff --git a/problems/0127.单词接龙.md b/problems/0127.单词接龙.md index f1c6f182..6d719691 100644 --- a/problems/0127.单词接龙.md +++ b/problems/0127.单词接龙.md @@ -158,6 +158,57 @@ class Solution: return 0 ``` ## Go +```go +func ladderLength(beginWord string, endWord string, wordList []string) int { + wordMap, que, depth := getWordMap(wordList, beginWord), []string{beginWord}, 0 + for len(que) > 0 { + depth++ + qLen := len(que) // 单词的长度 + for i := 0; i < qLen; i++ { + word := que[0] + que = que[1:] // 首位单词出队 + candidates := getCandidates(word) + for _, candidate := range candidates { + if _, exist := wordMap[candidate]; exist { // 用生成的结果集去查询 + if candidate == endWord { + return depth + 1 + } + delete(wordMap, candidate) // 删除集合中的用过的结果 + que = append(que, candidate) + } + } + } + } + return 0 +} + + +// 获取单词Map为后续的查询增加速度 +func getWordMap(wordList []string, beginWord string) map[string]int { + wordMap := make(map[string]int) + for i, word := range wordList { + if _, exist := wordMap[word]; !exist { + if word != beginWord { + wordMap[word] = i + } + } + } + return wordMap +} + +// 用26个英文字母分别替换掉各个位置的字母，生成一个结果集 +func getCandidates(word string) []string { + var res []string + for i := 0; i < 26; i++ { + for j := 0; j < len(word); j++ { + if word[j] != byte(int('a')+i) { + res = append(res, word[:j]+string(int('a')+i)+word[j+1:]) + } + } + } + return res +} +``` ## JavaScript ```javascript From bc8030b45a7e2a6cdefe8944cb624d4fa1f7d1a1 Mon Sep 17 00:00:00 2001 From: = <=> Date: Sun, 28 Aug 2022 14:31:18 +0800 Subject: [PATCH 27/32] =?UTF-8?q?Update=200347.=E5=89=8DK=E4=B8=AA?= =?UTF-8?q?=E9=AB=98=E9=A2=91=E5=85=83=E7=B4=A0.md=EF=BC=8C=E6=96=B0?= =?UTF-8?q?=E5=A2=9EComparator=E6=8E=A5=E5=8F=A3=E8=AF=B4=E6=98=8E?= =?UTF-8?q?=E5=8F=8A=E5=A4=A7=E9=A1=B6=E5=A0=86=E3=80=81=E5=B0=8F=E9=A1=B6?= =?UTF-8?q?=E5=A0=86=E4=B8=A4=E7=A7=8Djava=E5=AE=9E=E7=8E=B0?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0347.前K个高频元素.md | 59 +++++++++++++++++++++------- 1 file changed, 45 insertions(+), 14 deletions(-) diff --git a/problems/0347.前K个高频元素.md b/problems/0347.前K个高频元素.md index d4059b9b..8256e629 100644 --- a/problems/0347.前K个高频元素.md +++ b/problems/0347.前K个高频元素.md @@ -140,24 +140,55 @@ public: Java： ```java +/*Comparator接口说明: + * 返回负数，形参中第一个参数排在前面；返回正数，形参中第二个参数排在前面 + * 对于队列：排在前面意味着往队头靠 + * 对于堆（使用PriorityQueue实现）：从队头到队尾按从小到大排就是最小堆（小顶堆）， + * 从队头到队尾按从大到小排就是最大堆（大顶堆）--->队头元素相当于堆的根节点 + * */ class Solution { - public int[] topKFrequent(int[] nums, int k) { - int[] result = new int[k]; - HashMap map = new HashMap<>(); - for (int num : nums) { - map.put(num, map.getOrDefault(num, 0) + 1); + //解法1：基于大顶堆实现 + public int[] topKFrequent1(int[] nums, int k) { + Map map = new HashMap<>();//key为数组元素值,val为对应出现次数 + for(int num:nums){ + map.put(num,map.getOrDefault(num,0)+1); } - - Set> entries = map.entrySet(); - // 根据map的value值，构建于一个大顶堆（o1 - o2: 小顶堆， o2 - o1 : 大顶堆） - PriorityQueue> queue = new PriorityQueue<>((o1, o2) -> o2.getValue() - o1.getValue()); - for (Map.Entry entry : entries) { - queue.offer(entry); + //在优先队列中存储二元组(num,cnt),cnt表示元素值num在数组中的出现次数 + //出现次数按从队头到队尾的顺序是从大到小排,出现次数最多的在队头(相当于大顶堆) + PriorityQueue pq = new PriorityQueue<>((pair1, pair2)->pair2[1]-pair1[1]); + for(Map.Entry entry:map.entrySet()){//大顶堆需要对所有元素进行排序 + pq.add(new int[]{entry.getKey(),entry.getValue()}); } - for (int i = k - 1; i >= 0; i--) { - result[i] = queue.poll().getKey(); + int[] ans = new int[k]; + for(int i=0;i map = new HashMap<>();//key为数组元素值,val为对应出现次数 + for(int num:nums){ + map.put(num,map.getOrDefault(num,0)+1); + } + //在优先队列中存储二元组(num,cnt),cnt表示元素值num在数组中的出现次数 + //出现次数按从队头到队尾的顺序是从小到大排,出现次数最低的在队头(相当于小顶堆) + PriorityQueue pq = new PriorityQueue<>((pair1,pair2)->pair1[1]-pair2[1]); + for(Map.Entry entry:map.entrySet()){//小顶堆只需要维持k个元素有序 + if(pq.size()pq.peek()[1]){//当前元素出现次数大于小顶堆的根结点(这k个元素中出现次数最少的那个) + pq.poll();//弹出队头(小顶堆的根结点),即把堆里出现次数最少的那个删除,留下的就是出现次数多的了 + pq.add(new int[]{entry.getKey(),entry.getValue()}); + } + } + } + int[] ans = new int[k]; + for(int i=k-1;i>=0;i--){//依次弹出小顶堆,先弹出的是堆的根,出现次数少,后面弹出的出现次数多 + ans[i] = pq.poll()[0]; + } + return ans; } } ``` From 39f80d5bc38868f70a85d865bba9a20fc4965cf3 Mon Sep 17 00:00:00 2001 From: Xin-Sheng-5075 <2061461846@qq.com> Date: Sun, 28 Aug 2022 19:13:24 +0800 Subject: [PATCH 28/32] Signed-off-by: Xin-Sheng-5075 <2061461846@qq.com> MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 添加题目 200.岛屿数量 的Java版本 --- problems/0200.岛屿数量.md | 32 ++++++++++++++++++++++++++++++++ 1 file changed, 32 insertions(+) diff --git a/problems/0200.岛屿数量.md b/problems/0200.岛屿数量.md index b88e5fd2..59c4e566 100644 --- a/problems/0200.岛屿数量.md +++ b/problems/0200.岛屿数量.md @@ -247,3 +247,35 @@ public: ## 其他语言版本 + +### Java + +下面的代码使用的是深度优先搜索 DFS 的做法。为了统计岛屿数量同时不重复记录，每当我们搜索到一个岛后，就将这个岛 “淹没” —— 将这个岛所占的地方从 “1” 改为 “0”，这样就不用担心后续会重复记录这个岛屿了。而 DFS 的过程就体现在 “淹没” 这一步中。详见代码： + +```java +public int numIslands(char[][] grid) { + int res = 0; //记录找到的岛屿数量 + for(int i = 0;i < grid.length;i++){ + for(int j = 0;j < grid[0].length;j++){ + //找到“1”，res加一，同时淹没这个岛 + if(grid[i][j] == '1'){ + res++; + dfs(grid,i,j); + } + } + } + return res; +} +//使用DFS“淹没”岛屿 +public void dfs(char[][] grid, int i, int j){ + //搜索边界：索引越界或遍历到了"0" + if(i < 0 || i >= grid.length || j < 0 || j >= grid[0].length || grid[i][j] == '0') return; + //将这块土地标记为"0" + grid[i][j] = '0'; + //根据"每座岛屿只能由水平方向或竖直方向上相邻的陆地连接形成"，对上下左右的相邻顶点进行dfs + dfs(grid,i - 1,j); + dfs(grid,i + 1,j); + dfs(grid,i,j + 1); + dfs(grid,i,j - 1); +} +``` \ No newline at end of file From d91233d0443753b747d139e00b081fa4f4c39fb1 Mon Sep 17 00:00:00 2001 From: huaqi <2931246464@qq.com> Date: Tue, 30 Aug 2022 21:18:16 +0800 Subject: [PATCH 29/32] =?UTF-8?q?0977.=E6=9C=89=E5=BA=8F=E6=95=B0=E7=BB=84?= =?UTF-8?q?=E7=9A=84=E5=B9=B3=E6=96=B9-typescript=E7=89=88=E6=9C=AC-?= =?UTF-8?q?=E4=BC=98=E5=8C=96=E5=86=99=E6=B3=95?= 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 | 22 ++++++++++++++-------- 1 file changed, 14 insertions(+), 8 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 eb9f42b1..501b9a42 100644 --- a/problems/0977.有序数组的平方.md +++ b/problems/0977.有序数组的平方.md @@ -239,18 +239,24 @@ Typescript： ```typescript function sortedSquares(nums: number[]): number[] { - let left: number = 0, right: number = nums.length - 1; - let resArr: number[] = new Array(nums.length); - let resArrIndex: number = resArr.length - 1; + const ans: number[] = []; + let left = 0, + right = nums.length - 1; + while (left <= right) { - if (Math.abs(nums[left]) < Math.abs(nums[right])) { - resArr[resArrIndex] = nums[right--] ** 2; + // 右侧的元素不需要取绝对值，nums 为非递减排序的整数数组 + // 在同为负数的情况下，左侧的平方值一定大于右侧的平方值 + if (Math.abs(nums[left]) > nums[right]) { + // 使用 Array.prototype.unshift() 直接在数组的首项插入当前最大值 + ans.unshift(nums[left] ** 2); + left++; } else { - resArr[resArrIndex] = nums[left++] ** 2; + ans.unshift(nums[right] ** 2); + right--; } - resArrIndex--; } - return resArr; + + return ans; }; ``` From 3f96750bece9167e25e52a61c1f52cdcdecdb8ea Mon Sep 17 00:00:00 2001 From: Wen Date: Tue, 30 Aug 2022 22:30:25 +0800 Subject: [PATCH 30/32] =?UTF-8?q?=E4=BC=98=E5=8C=96=20242.=20=E6=9C=89?= =?UTF-8?q?=E6=95=88=E7=9A=84=E5=AD=97=E6=AF=8D=E5=BC=82=E4=BD=8D=E8=AF=8D?= =?UTF-8?q?=20Java=E8=A7=A3=E6=B3=95?= 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 | 20 +++++++++++++------- 1 file changed, 13 insertions(+), 7 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 87302482..d54285eb 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 @@ -92,18 +92,24 @@ public: Java： ```java +/** + * 242. 有效的字母异位词 字典解法 + * 时间复杂度O(m+n) 空间复杂度O(1) + */ class Solution { public boolean isAnagram(String s, String t) { - int[] record = new int[26]; - for (char c : s.toCharArray()) { - record[c - 'a'] += 1; + + for (int i = 0; i < s.length(); i++) { + record[s.charAt(i) - 'a']++; } - for (char c : t.toCharArray()) { - record[c - 'a'] -= 1; + + for (int i = 0; i < t.length(); i++) { + record[t.charAt(i) - 'a']--; } - for (int i : record) { - if (i != 0) { + + for (int count: record) { + if (count != 0) { return false; } } From a275c2cc553ad35e50d736e0630947a5bdf2d13e Mon Sep 17 00:00:00 2001 From: YukunJ Date: Tue, 30 Aug 2022 21:51:30 -0400 Subject: [PATCH 31/32] =?UTF-8?q?Update:=20=E6=B3=A8=E9=87=8A=E5=A2=9E?= =?UTF-8?q?=E5=8A=A0=E4=BD=BF=E7=94=A8STL=E6=B1=82=E5=92=8C=E7=9A=84?= =?UTF-8?q?=E7=AE=80=E4=BE=BF=E5=B0=8F=E8=B4=B4=E5=A3=AB?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0416.分割等和子集.md | 9 +++++++-- 1 file changed, 7 insertions(+), 2 deletions(-) diff --git a/problems/0416.分割等和子集.md b/problems/0416.分割等和子集.md index 9c90ea27..fa677da0 100644 --- a/problems/0416.分割等和子集.md +++ b/problems/0416.分割等和子集.md @@ -143,12 +143,17 @@ dp[j]的数值一定是小于等于j的。 class Solution { public: bool canPartition(vector& nums) { - // 使用标准库函数 便捷求和 - int sum = accumulate(nums.begin(), nums.end(), 0); + int sum = 0; + // dp[i]中的i表示背包内总和 // 题目中说：每个数组中的元素不会超过 100，数组的大小不会超过 200 // 总和不会大于20000，背包最大只需要其中一半，所以10001大小就可以了 vector dp(10001, 0); + for (int i = 0; i < nums.size(); i++) { + sum += nums[i]; + } + // 也可以使用库函数一步求和 + // int sum = accumulate(nums.begin(), nums.end(), 0); if (sum % 2 == 1) return false; int target = sum / 2; From a8d3874f550794df371ecb762cb4e391808e75b3 Mon Sep 17 00:00:00 2001 From: Yuki Chen <826992207@qq.com> Date: Wed, 31 Aug 2022 16:44:44 +0800 Subject: [PATCH 32/32] =?UTF-8?q?=E6=B7=BB=E5=8A=A00053.=E6=9C=80=E5=A4=A7?= =?UTF-8?q?=E5=AD=90=E5=BA=8F=E5=92=8C=EF=BC=88=E5=8A=A8=E6=80=81=E8=A7=84?= =?UTF-8?q?=E5=88=92=EF=BC=89=E7=A9=BA=E9=97=B4=E5=A4=8D=E6=9D=82=E5=BA=A6?= =?UTF-8?q?=E4=B8=BAO(1)=E7=9A=84=E8=A7=A3=E6=B3=95=20Java=E7=89=88?= =?UTF-8?q?=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0053.最大子序和（动态规划）.md | 14 ++++++++++++++ 1 file changed, 14 insertions(+) diff --git a/problems/0053.最大子序和（动态规划）.md b/problems/0053.最大子序和（动态规划）.md index 00f3eb84..23d23400 100644 --- a/problems/0053.最大子序和（动态规划）.md +++ b/problems/0053.最大子序和（动态规划）.md @@ -120,6 +120,20 @@ Java： return res; } ``` +```Java +//因为dp[i]的递推公式只与前一个值有关，所以可以用一个变量代替dp数组，空间复杂度为O(1) +class Solution { + public int maxSubArray(int[] nums) { + int res = nums[0]; + int pre = nums[0]; + for(int i = 1; i < nums.length; i++) { + pre = Math.max(pre + nums[i], nums[i]); + res = Math.max(res, pre); + } + return res; + } +} +``` Python： ```python