Merge branch 'master' of github.com:youngyangyang04/leetcode-master

problems/0015.三数之和.md

@@ -435,5 +435,79 @@ func threeSum(_ nums: [Int]) -> [[Int]] {
 }
 ```
 
+C:
+```C
+//qsort辅助cmp函数
+int cmp(const void* ptr1, const void* ptr2) {
+    return *((int*)ptr1) > *((int*)ptr2);
+}
+
+int** threeSum(int* nums, int numsSize, int* returnSize, int** returnColumnSizes) {
+    //开辟ans数组空间
+    int **ans = (int**)malloc(sizeof(int*) * 18000);
+    int ansTop = 0;
+    //若传入nums数组大小小于3，则需要返回数组大小为0
+    if(numsSize < 3) {
+        *returnSize = 0;
+        return ans;
+    }
+    //对nums数组进行排序
+    qsort(nums, numsSize, sizeof(int), cmp);
+    
+
+    int i;
+    //用for循环遍历数组，结束条件为i < numsSize - 2(因为要预留左右指针的位置)
+    for(i = 0; i < numsSize - 2; i++) {
+        //若当前i指向元素>0，则代表left和right以及i的和大于0。直接break
+        if(nums[i] > 0)
+            break;
+        //去重：i > 0 && nums[i] == nums[i-1]
+        if(i > 0 && nums[i] == nums[i-1])
+            continue;
+        //定义左指针和右指针
+        int left = i + 1;
+        int right = numsSize - 1;
+        //当右指针比左指针大时进行循环
+        while(right > left) {
+            //求出三数之和
+            int sum = nums[right] + nums[left] + nums[i];
+            //若和小于0，则左指针+1（因为左指针右边的数比当前所指元素大）
+            if(sum < 0)
+                left++;
+            //若和大于0，则将右指针-1
+            else if(sum > 0)
+                right--;
+            //若和等于0
+            else {
+                //开辟一个大小为3的数组空间，存入nums[i], nums[left]和nums[right]
+                int* arr = (int*)malloc(sizeof(int) * 3);
+                arr[0] = nums[i];
+                arr[1] = nums[left];
+                arr[2] = nums[right];
+                //将开辟数组存入ans中
+                ans[ansTop++] = arr;
+                //去重
+                while(right > left && nums[right] == nums[right - 1])
+                    right--;
+                while(left < right && nums[left] == nums[left + 1])
+                    left++;
+                //更新左右指针
+                left++;
+                right--;
+            }
+        }
+    }
+
+    //设定返回的数组大小
+    *returnSize = ansTop;
+    *returnColumnSizes = (int*)malloc(sizeof(int) * ansTop);
+    int z;
+    for(z = 0; z < ansTop; z++) {
+        (*returnColumnSizes)[z] = 3;
+    }
+    return ans;
+}
+```
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0019.删除链表的倒数第N个节点.md

@@ -173,8 +173,7 @@ var removeNthFromEnd = function(head, n) {
     let ret = new ListNode(0, head),
         slow = fast = ret;
     while(n--) fast = fast.next;
-    if(!fast) return ret.next;
+    while (fast.next !== null) {
-    while (fast.next) {
         fast = fast.next; 
         slow = slow.next
     };

problems/0040.组合总和II.md

@@ -457,6 +457,37 @@ var combinationSum2 = function(candidates, target) {
     }
 };
 ```
+**使用used去重**
+```js
+var combinationSum2 = function(candidates, target) {
+    let res = [];
+    let path = [];
+    let total = 0;
+    const len = candidates.length;
+    candidates.sort((a, b) => a - b);
+    let used = new Array(len).fill(false);
+    const backtracking = (startIndex) => {
+        if (total === target) {
+            res.push([...path]);
+            return;
+        }
+        for(let i = startIndex; i < len && total < target; i++) {
+            const cur = candidates[i];
+            if (cur > target - total || (i > 0 && cur === candidates[i - 1] && !used[i - 1])) continue;
+            path.push(cur);
+            total += cur;
+            used[i] = true;
+            backtracking(i + 1);
+            path.pop();
+            total -= cur;
+            used[i] = false;
+        }
+    }
+    backtracking(0);
+    return res;
+};
+```
+
 ## C 
 
 ```c

problems/0098.验证二叉搜索树.md

@@ -414,19 +414,24 @@ class Solution:
 import "math"
 
 func isValidBST(root *TreeNode) bool {
+	// 二叉搜索树也可以是空树
     if root == nil {
         return true
     }
-	return isBST(root, math.MinInt64, math.MaxFloat64)
+    // 由题目中的数据限制可以得出min和max
+    return check(root,math.MinInt64,math.MaxInt64)
 }
-func isBST(root *TreeNode, min, max int) bool {
+
-	if root == nil {
+func check(node *TreeNode,min,max int64) bool {
+    if node == nil {
         return true
     }
-	if min >= root.Val || max <= root.Val {
+
+    if min >= int64(node.Val) || max <= int64(node.Val) {
         return false
     }
-	return isBST(root.Left, min, root.Val) && isBST(root.Right, root.Val, max)
+    // 分别对左子树和右子树递归判断，如果左子树和右子树都符合则返回true
+    return check(node.Right,int64(node.Val),max) && check(node.Left,min,int64(node.Val))
 }
 ```
 ```go

problems/0150.逆波兰表达式求值.md

@@ -241,6 +241,38 @@ class Solution:
 
 ```
 
+Swift：
+```Swift
+func evalRPN(_ tokens: [String]) -> Int {
+    var stack = [Int]()
+    for c in tokens {
+        let v = Int(c)
+        if let num = v {
+            // 遇到数字直接入栈
+            stack.append(num)
+        } else {
+            // 遇到运算符, 取出栈顶两元素计算, 结果压栈
+            var res: Int = 0
+            let num2 = stack.popLast()!
+            let num1 = stack.popLast()!
+            switch c {
+            case "+":
+                res = num1 + num2
+            case "-":
+                res = num1 - num2
+            case "*":
+                res = num1 * num2
+            case "/":
+                res = num1 / num2
+            default:
+                break
+            }
+            stack.append(res)
+        }
+    }
+    return stack.last!
+}
+```
 
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 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0205.同构字符串.md

@@ -90,6 +90,24 @@ class Solution {
 ## Python
 
 ```python
+class Solution:
+    def isIsomorphic(self, s: str, t: str) -> bool:
+        default_dict1 = defaultdict(str)
+        default_dict2 = defaultdict(str)
+    
+        if len(s) != len(t): return false
+
+        for i in range(len(s)):
+            if not default_dict1[s[i]]:
+                default_dict1[s[i]] = t[i]
+            
+            if not default_dict2[t[i]]:
+                default_dict2[t[i]] = s[i]
+
+            if default_dict1[s[i]] != t[i] or default_dict2[t[i]] != s[i]:
+                return False
+            
+        return True
 ```
 
 ## Go

problems/0236.二叉树的最近公共祖先.md

@@ -221,42 +221,30 @@ public:
 
 
 ## Java 
+
 ```Java
 class Solution {
     public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
-        return lowestCommonAncestor1(root, p, q);
+        if (root == null || root == p || root == q) { // 递归结束条件
-    }
-    public TreeNode lowestCommonAncestor1(TreeNode root, TreeNode p, TreeNode q) {
-        if (root == null || root == p || root == q) {
             return root;
         }
-        TreeNode left = lowestCommonAncestor1(root.left, p, q);
-        TreeNode right = lowestCommonAncestor1(root.right, p, q);
-        if (left != null && right != null) {// 左右子树分别找到了，说明此时的root就是要求的结果
-            return root;
-        }
-        if (left == null) {
-            return right;
-        }
-        return left;
-    }
-}
-```
 
-
+        // 后序遍历
-```java
-// 代码精简版
-class Solution {
-    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
-        if (root == null || root.val == p.val ||root.val == q.val) return root;
         TreeNode left = lowestCommonAncestor(root.left, p, q);
         TreeNode right = lowestCommonAncestor(root.right, p, q);
-        if (left != null && right != null) return root;
+
-        else if (left == null && right != null) return right;
+        if(left == null && right == null) { // 若未找到节点 p 或 q
-        else if (left != null && right == null) return left;
+            return null;
-        else return null;
+        }else if(left == null && right != null) { // 若找到一个节点
+            return right;
+        }else if(left != null && right == null) { // 若找到一个节点
+            return left;
+        }else { // 若找到两个节点
+            return root;
         }
     }
+}
+
 ```
 
 ## Python 

problems/0239.滑动窗口最大值.md

@@ -418,6 +418,112 @@ var maxSlidingWindow = function (nums, k) {
 };
 ```
 
+Swift:
+```Swift
+/// 双向链表
+class DoublyListNode {
+    var head: DoublyListNode?
+    var tail: DoublyListNode?
+    var next: DoublyListNode?
+    var pre: DoublyListNode?
+    var value: Int = 0
+    init(_ value: Int = 0) {
+        self.value = value
+    }
+    func isEmpty() -> Bool {
+        return self.head == nil
+    }
+    func first() -> Int? {
+        return self.head?.value
+    }
+    func last() -> Int? {
+        return self.tail?.value
+    }
+    func removeFirst() {
+        if isEmpty() {
+            return
+        }
+        let next = self.head!.next
+        self.head?.next = nil// 移除首节点
+        next?.pre = nil
+        self.head = next
+    }
+    func removeLast() {
+        if let tail = self.tail {
+            if let pre = tail.pre {
+                self.tail?.pre = nil
+                pre.next = nil
+                self.tail = pre
+            } else {
+                self.head = nil
+                self.tail = nil
+            }
+        }
+    }
+    func append(_ value: Int) {
+        let node = DoublyListNode(value)
+        if self.head != nil {
+            node.pre = self.tail
+            self.tail?.next = node
+            self.tail = node
+        } else {
+            self.head = node
+            self.tail = node
+            self.pre = nil
+            self.next = nil
+        }
+    }
+}
+// 单调队列, 从大到小
+class MyQueue {
+//    var queue: [Int]!// 用数组会超时
+    var queue: DoublyListNode!
+    init() {
+//        queue = [Int]()
+        queue = DoublyListNode()
+    }
+    // 滑动窗口时弹出第一个元素, 如果相等再弹出
+    func pop(x: Int) {
+        if !queue.isEmpty() && front() == x {
+            queue.removeFirst()
+        }
+    }
+    // 滑动窗口时添加下一个元素, 移除队尾比 x 小的元素 始终保证队头 > 队尾
+    func push(x: Int) {
+        while !queue.isEmpty() && queue.last()! < x {
+            queue.removeLast()
+        }
+        queue.append(x)
+    }
+    // 此时队头就是滑动窗口最大值
+    func front() -> Int {
+        return queue.first() ?? -1
+    }
+}
+
+class Solution {
+    func maxSlidingWindow(_ nums: [Int], _ k: Int) -> [Int] {
+        // 存放结果
+        var res = [Int]()
+        let queue = MyQueue()
+        // 先将前K个元素放入队列
+        for i in 0 ..< k {
+            queue.push(x: nums[i])
+        }
+        // 添加当前队列最大值到结果数组
+        res.append(queue.front())
+        for i in k ..< nums.count {
+            // 滑动窗口移除最前面元素
+            queue.pop(x: nums[i - k])
+            // 滑动窗口添加下一个元素
+            queue.push(x: nums[i])
+            // 保存当前队列最大值
+            res.append(queue.front())
+        }
+        return res
+    }
+}
+```
 
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0416.分割等和子集.md

@@ -68,7 +68,7 @@
 
 * 背包的体积为sum / 2
 * 背包要放入的商品（集合里的元素）重量为 元素的数值，价值也为元素的数值
-* 背包如何正好装满，说明找到了总和为 sum / 2 的子集。
+* 背包如果正好装满，说明找到了总和为 sum / 2 的子集。
 * 背包中每一个元素是不可重复放入。
 
 以上分析完，我们就可以套用01背包，来解决这个问题了。

problems/0701.二叉搜索树中的插入操作.md

@@ -236,16 +236,13 @@ class Solution {
 ```java
 class Solution {
     public TreeNode insertIntoBST(TreeNode root, int val) {
-        return buildTree(root, val);
-    }
-
-    public TreeNode buildTree(TreeNode root, int val){
         if (root == null) // 如果当前节点为空，也就意味着val找到了合适的位置，此时创建节点直接返回。
             return new TreeNode(val);
+            
         if (root.val < val){
-            root.right = buildTree(root.right, val); // 递归创建右子树
+            root.right = insertIntoBST(root.right, val); // 递归创建右子树
         }else if (root.val > val){
-            root.left = buildTree(root.left, val); // 递归创建左子树
+            root.left = insertIntoBST(root.left, val); // 递归创建左子树
         }
         return root;
     }

problems/0704.二分查找.md

@@ -276,11 +276,9 @@ func search(nums []int, target int) int {
 ```
 
 **JavaScript:**
+（版本一）左闭右闭区间
 
 ```js
-
-// （版本一）左闭右闭区间
-
 /**
  * @param {number[]} nums
  * @param {number} target
@@ -302,9 +300,10 @@ var search = function(nums, target) {
     }
     return -1;
 };
+```
+（版本二）左闭右开区间
 
-// （版本二）左闭右开区间
+```js
-
 /**
  * @param {number[]} nums
  * @param {number} target
@@ -325,7 +324,6 @@ var search = function(nums, target) {
     }
     return -1;
 };
-
 ```
 
 **Ruby:**

problems/0977.有序数组的平方.md

@@ -199,27 +199,26 @@ impl Solution {
 Javascript：
 ```Javascript
 /**
- * @desc two pointers solution
+ * @param {number[]} nums
- * @link https://leetcode-cn.com/problems/squares-of-a-sorted-array/
+ * @return {number[]}
- * @param nums Array  e.g. [-4,-1,0,3,10]
- * @return {array}    e.g. [0,1,9,16,100]
  */
-const sortedSquares = function (nums) {
+var sortedSquares = function(nums) {
-    let res = []
+    let n = nums.length;
-    for (let i = 0, j = nums.length - 1; i <= j;) {
+    let res = new Array(n).fill(0);
-      const left = Math.abs(nums[i])
+    let i = 0, j = n - 1, k = n - 1;
-      const right = Math.abs(nums[j])
+    while (i <= j) {
-      if (right > left) {
+        let left = nums[i] * nums[i],
-        // push element to the front of the array
+            right = nums[j] * nums[j];
-        res.unshift(right * right)
+        if (left < right) {
-        j--
+            res[k--] = right;
+            j--;
         } else {
-        res.unshift(left * left)
+            res[k--] = left;
-        i++
+            i++;
         }
     }
-    return res
+    return res;
-  }
+};
 ```
 
 Swift: