Merge branch 'youngyangyang04:master' into master

problems/0005.最长回文子串.md

@@ -263,21 +263,38 @@ public:
 Java：
 
 ```java
-public int[] twoSum(int[] nums, int target) {
+// 双指针 动态规划
-    int[] res = new int[2];
+class Solution {
-    if(nums == null || nums.length == 0){
+    public String longestPalindrome(String s) {
-        return res;
+        if (s.length() == 0 || s.length() == 1) return s;
-    }
+        int length = 1;
-    Map<Integer, Integer> map = new HashMap<>();
+        int index = 0;
-    for(int i = 0; i < nums.length; i++){
+        boolean[][] palindrome = new boolean[s.length()][s.length()];
-        int temp = target - nums[i];
+        for (int i = 0; i < s.length(); i++) {
-        if(map.containsKey(temp)){
+            palindrome[i][i] = true;
-            res[1] = i;
-            res[0] = map.get(temp);
         }
-        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;
 }
 ```
 

problems/0051.N皇后.md

@@ -222,56 +222,6 @@ 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
@@ -341,6 +291,55 @@ 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
@@ -396,6 +395,8 @@ func isValid(n, row, col int, chessboard [][]string) bool {
 	return true
 }
 ```
+
+
 ### Javascript 
 ```Javascript
 var solveNQueens = function(n) {

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 dp = new Array(n + 1).fill(0);
-    const weight = [1,2];
+    const m = 2;
     dp[0] = 1;
-    for(let i = 0; i <= n; i++){ //先遍历背包
+    for(let i = 1; i <= n; i++){
-        for(let j = 0; j < weight.length; j++){ // 再遍历物品
+        for(let j = 1; j <= m; j++){
-            if(i >= weight[j]) dp[i] += dp[i-weight[j]];
+            if(i >= j) {
+	    	dp[i] += dp[i - j];
+	     }
         }
     }
     return dp[n];

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<vector<int>> 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<vector<int>> result;

problems/0332.重新安排行程.md

@@ -262,6 +262,43 @@ for (pair<string, int>target : targets[result[result.size() - 1]])
 
 ### java 
 	
+```java
+class Solution {
+    private LinkedList<String> res;
+    private LinkedList<String> path = new LinkedList<>();
+
+    public List<String> findItinerary(List<List<String>> 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<List<String>> 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 {
     private Deque<String> res;

problems/0347.前K个高频元素.md

@@ -268,99 +268,102 @@ func topKFrequent(nums []int, k int) []int {
 
 
 
-javaScript:
+JavaScript:
 ```js
-/**
+// js 没有堆 需要自己构造
- * @param {number[]} nums
+class Heap {
- * @param {number} k
+    constructor(compareFn) {
- * @return {number[]}
+        this.compareFn = compareFn;
- */
+        this.queue = [];
-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();
     }
-  }
 
-  const ret = [];
+    // 添加
+    push(item) {
+        // 推入元素
+        this.queue.push(item);
 
-  for(let i = priorityQueue.size() - 1; i >= 0; i--) {
+        // 上浮
-    ret[i] = priorityQueue.pop()[0];
+        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：

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：

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。
 

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;