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

problems/0015.三数之和.md

@@ -434,6 +434,46 @@ class Solution {
 }
 ```
 
+Swift:
+```swift
+// 双指针法
+func threeSum(_ nums: [Int]) -> [[Int]] {
+    var res = [[Int]]()
+    var sorted = nums
+    sorted.sort()
+    for i in 0 ..< sorted.count {
+        if sorted[i] > 0 {
+            return res
+        }
+        if i > 0 && sorted[i] == sorted[i - 1] {
+            continue
+        }
+        var left = i + 1
+        var right = sorted.count - 1
+        while left < right {
+            let sum = sorted[i] + sorted[left] + sorted[right]
+            if sum < 0 {
+                left += 1
+            } else if sum > 0 {
+                right -= 1
+            } else {
+                res.append([sorted[i], sorted[left], sorted[right]])
+
+                while left < right && sorted[left] == sorted[left + 1] {
+                    left += 1
+                }
+                while left < right && sorted[right] == sorted[right - 1] {
+                    right -= 1
+                }
+
+                left += 1
+                right -= 1
+            }
+        }
+    }
+    return res
+}
+```
 
 -----------------------
 * 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)

problems/0018.四数之和.md

@@ -354,6 +354,54 @@ class Solution {
 }
 ```
 
+Swift:
+```swift
+func fourSum(_ nums: [Int], _ target: Int) -> [[Int]] {
+    var res = [[Int]]()
+    var sorted = nums
+    sorted.sort()
+    for k in 0 ..< sorted.count {
+        // 这种剪枝不行,target可能是负数
+//            if sorted[k] > target {
+//                return res
+//            }
+        // 去重
+        if k > 0 && sorted[k] == sorted[k - 1] {
+            continue
+        }
+
+        let target2 = target - sorted[k]
+        for i in (k + 1) ..< sorted.count {
+            if i > (k + 1) && sorted[i] == sorted[i - 1] {
+                continue
+            }
+            var left = i + 1
+            var right = sorted.count - 1
+            while left < right {
+                let sum = sorted[i] + sorted[left] + sorted[right]
+                if sum < target2 {
+                    left += 1
+                } else if sum > target2 {
+                    right -= 1
+                } else {
+                    res.append([sorted[k], sorted[i], sorted[left], sorted[right]])
+                    while left < right && sorted[left] == sorted[left + 1] {
+                        left += 1
+                    }
+                    while left < right && sorted[right] == sorted[right - 1]  {
+                        right -= 1
+                    }
+                    // 找到答案 双指针同时收缩
+                    left += 1
+                    right -= 1
+                }
+            }
+        }
+    }
+    return res
+}
+```
+
 -----------------------
 * 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
 * B站视频：[代码随想录](https://space.bilibili.com/525438321)

problems/0042.接雨水.md

@@ -391,6 +391,7 @@ class Solution {
     }
 }
 ```
+
 动态规划法
 ```java
 class Solution {
@@ -418,6 +419,54 @@ class Solution {
     }
 }
 ```
+
+单调栈法
+```java
+class Solution {
+    public int trap(int[] height){
+        int size = height.length;
+
+        if (size <= 2) return 0;
+
+        // in the stack, we push the index of array
+        // using height[] to access the real height
+        Stack<Integer> stack = new Stack<Integer>();
+        stack.push(0);
+
+        int sum = 0;
+        for (int index = 1; index < size; index++){
+            int stackTop = stack.peek();
+            if (height[index] < height[stackTop]){
+                stack.push(index);
+            }else if (height[index] == height[stackTop]){
+                // 因为相等的相邻墙，左边一个是不可能存放雨水的，所以pop左边的index, push当前的index
+                stack.pop();
+                stack.push(index);
+            }else{
+                //pop up all lower value
+                int heightAtIdx = height[index];
+                while (!stack.isEmpty() && (heightAtIdx > height[stackTop])){
+                    int mid = stack.pop();
+
+                    if (!stack.isEmpty()){
+                        int left = stack.peek();
+
+                        int h = Math.min(height[left], height[index]) - height[mid];
+                        int w = index - left - 1;
+                        int hold = h * w;
+                        if (hold > 0) sum += hold;
+                        stackTop = stack.peek();
+                    }
+                }
+                stack.push(index);
+            }
+        }
+
+        return sum;
+    }
+}
+```
+
 Python:
 
 双指针法

problems/0084.柱状图中最大的矩形.md

@@ -228,6 +228,50 @@ class Solution {
 }
 ```
 
+单调栈
+```java
+class Solution {
+    int largestRectangleArea(int[] heights) {
+        Stack<Integer> st = new Stack<Integer>();
+
+        // 数组扩容，在头和尾各加入一个元素
+        int [] newHeights = new int[heights.length + 2];
+        newHeights[0] = 0;
+        newHeights[newHeights.length - 1] = 0;
+        for (int index = 0; index < heights.length; index++){
+            newHeights[index + 1] = heights[index];
+        }
+
+        heights = newHeights;
+
+        st.push(0);
+        int result = 0;
+        // 第一个元素已经入栈，从下表1开始
+        for (int i = 1; i < heights.length; i++) {
+            // 注意heights[i] 是和heights[st.top()] 比较 ，st.top()是下表
+            if (heights[i] > heights[st.peek()]) {
+                st.push(i);
+            } else if (heights[i] == heights[st.peek()]) {
+                st.pop(); // 这个可以加，可以不加，效果一样，思路不同
+                st.push(i);
+            } else {
+                while (heights[i] < heights[st.peek()]) { // 注意是while
+                    int mid = st.peek();
+                    st.pop();
+                    int left = st.peek();
+                    int right = i;
+                    int w = right - left - 1;
+                    int h = heights[mid];
+                    result = Math.max(result, w * h);
+                }
+                st.push(i);
+            }
+        }
+        return result;
+    }
+}
+```
+
 Python:
 
 动态规划

problems/0106.从中序与后序遍历序列构造二叉树.md

@@ -607,6 +607,7 @@ class Solution {
         for (int i = inLeft; i < inRight; i++) {
             if (inorder[i] == rootVal) {
                 rootIndex = i;
+                break;
             }
         }
         // 根据rootIndex划分左右子树

problems/0112.路径总和.md

@@ -411,6 +411,31 @@ class solution {
 }
 ```
 
+```java
+// 解法2
+class Solution {
+    List<List<Integer>> result;
+    LinkedList<Integer> path;
+    public List<List<Integer>> pathSum (TreeNode root,int targetSum) {
+        result = new LinkedList<>();
+        path = new LinkedList<>();
+        travesal(root, targetSum);
+        return result;
+    }
+    private void travesal(TreeNode root,  int count) {
+        if (root == null) return;
+        path.offer(root.val);
+        count -= root.val;
+        if (root.left == null && root.right == null && count == 0) {
+            result.add(new LinkedList<>(path));
+        }
+        travesal(root.left, count);
+        travesal(root.right, count);
+        path.removeLast(); // 回溯
+    }
+}
+```
+
 ## python
 
 0112.路径总和

problems/0129.求根到叶子节点数字之和.md

@@ -164,6 +164,52 @@ public:
 
 Java： 
 
+```java
+class Solution {
+    List<Integer> path = new ArrayList<>();
+    int res = 0;
+    public int sumNumbers(TreeNode root) {
+        // 如果节点为0，那么就返回0
+        if (root == null) return 0;
+        // 首先将根节点放到集合中
+        path.add(root.val);
+        // 开始递归
+        recur(root);
+        return res;
+    }
+
+    public void recur(TreeNode root){
+        if (root.left == null && root.right == null) {
+            // 当是叶子节点的时候，开始处理
+            res += listToInt(path);
+            return;
+        }
+
+        if (root.left != null){
+            // 注意有回溯
+            path.add(root.left.val);
+            recur(root.left);
+            path.remove(path.size() - 1);
+        }
+        if (root.right != null){
+            // 注意有回溯
+            path.add(root.right.val);
+            recur(root.right);
+            path.remove(path.size() - 1);
+        }
+        return;
+    }
+    public int listToInt(List<Integer> path){
+        int sum = 0;
+        for (Integer num:path){
+            // sum * 10 表示进位
+            sum = sum * 10 + num;
+        }
+        return sum;
+    }
+}
+```
+
 Python：
 ```python3
 class Solution:

problems/0143.重排链表.md

@@ -177,6 +177,7 @@ public:
 Java：
 
 ```java
+// 方法三
 public class ReorderList {
     public void reorderList(ListNode head) {
         ListNode fast = head, slow = head;
@@ -219,6 +220,76 @@ public class ReorderList {
         return headNode.next;
     }
 }
+
+-------------------------------------------------------------------------
+// 方法一 Java实现，使用数组存储节点
+ class Solution {
+    public void reorderList(ListNode head) {
+        // 双指针的做法
+        ListNode cur = head;
+        // ArrayList底层是数组，可以使用下标随机访问
+        List<ListNode> list = new ArrayList<>();  
+        while (cur != null){
+            list.add(cur);
+            cur = cur.next;
+        }
+        cur = head;  // 重新回到头部
+        int l = 1, r = list.size() - 1;  // 注意左边是从1开始
+        int count = 0;
+        while (l <= r){
+            if (count % 2 == 0){
+                // 偶数
+                cur.next = list.get(r);
+                r--;
+            }else {
+                // 奇数
+                cur.next = list.get(l);
+                l++;
+            }
+            // 每一次指针都需要移动
+            cur = cur.next;
+            count++;
+        }
+        // 当是偶数的话，需要做额外处理
+        if (list.size() % 2== 0){
+            cur.next = list.get(l);
+            cur = cur.next;
+        }
+
+        // 注意结尾要结束一波
+        cur.next = null;
+    }
+}
+-------------------------------------------------------------------------
+// 方法二：使用双端队列，简化了数组的操作，代码相对于前者更简洁（避免一些边界条件）
+class Solution {
+    public void reorderList(ListNode head) {
+        // 使用双端队列的方法来解决
+        Deque<ListNode> de = new LinkedList<>();
+        // 这里是取head的下一个节点，head不需要再入队了，避免造成重复
+        ListNode cur = head.next;  
+        while (cur != null){
+            de.offer(cur);
+            cur = cur.next;
+        }
+        cur = head;  // 回到头部
+
+        int count = 0;
+        while (!de.isEmpty()){
+            if (count % 2 == 0){
+                // 偶数，取出队列右边尾部的值
+                cur.next = de.pollLast();
+            }else {
+                // 奇数，取出队列左边头部的值
+                cur.next = de.poll();
+            }
+            cur  = cur.next;
+            count++;
+        }
+        cur.next = null;
+    }
+}
+
 ```
 
 Python：