Merge branch 'master' into master

Author: youngyangyang04

problems/0019.删除链表的倒数第N个节点.md

@@ -181,7 +181,71 @@ var removeNthFromEnd = function(head, n) {
     return ret.next;
 };
 ```
+TypeScript:
+
+版本一（快慢指针法）：
+
+```typescript
+function removeNthFromEnd(head: ListNode | null, n: number): ListNode | null {
+    let newHead: ListNode | null = new ListNode(0, head);
+    let slowNode: ListNode | null = newHead,
+        fastNode: ListNode | null = newHead;
+    for (let i = 0; i < n; i++) {
+        fastNode = fastNode.next;
+    }
+    while (fastNode.next) {
+        fastNode = fastNode.next;
+        slowNode = slowNode.next;
+    }
+    slowNode.next = slowNode.next.next;
+    return newHead.next;
+};
+```
+
+版本二（计算节点总数法）：
+
+```typescript
+function removeNthFromEnd(head: ListNode | null, n: number): ListNode | null {
+    let curNode: ListNode | null = head;
+    let listSize: number = 0;
+    while (curNode) {
+        curNode = curNode.next;
+        listSize++;
+    }
+    if (listSize === n) {
+        head = head.next;
+    } else {
+        curNode = head;
+        for (let i = 0; i < listSize - n - 1; i++) {
+            curNode = curNode.next;
+        }
+        curNode.next = curNode.next.next;
+    }
+    return head;
+};
+```
+
+版本三（递归倒退n法）：
+
+```typescript
+function removeNthFromEnd(head: ListNode | null, n: number): ListNode | null {
+    let newHead: ListNode | null = new ListNode(0, head);
+    let cnt = 0;
+    function recur(node) {
+        if (node === null) return;
+        recur(node.next);
+        cnt++;
+        if (cnt === n + 1) {
+            node.next = node.next.next;
+        }
+    }
+    recur(newHead);
+    return newHead.next;
+};
+```
+
 Kotlin:
+
 ```Kotlin
 fun removeNthFromEnd(head: ListNode?, n: Int): ListNode? {
     val pre = ListNode(0).apply {

problems/0024.两两交换链表中的节点.md

@@ -250,6 +250,38 @@ var swapPairs = function (head) {
 };
 ```
 
+TypeScript：
+
+```typescript
+function swapPairs(head: ListNode | null): ListNode | null {
+    /**
+     * 初始状态：
+     * curNode -> node1 -> node2 -> tmepNode
+     * 转换过程：
+     * curNode -> node2
+     * curNode -> node2 -> node1
+     * curNode -> node2 -> node1 -> tempNode
+     * curNode = node1
+     */
+    let retNode: ListNode | null = new ListNode(0, head),
+        curNode: ListNode | null = retNode,
+        node1: ListNode | null = null,
+        node2: ListNode | null = null,
+        tempNode: ListNode | null = null;
+
+    while (curNode && curNode.next && curNode.next.next) {
+        node1 = curNode.next;
+        node2 = curNode.next.next;
+        tempNode = node2.next;
+        curNode.next = node2;
+        node2.next = node1;
+        node1.next = tempNode;
+        curNode = node1;
+    }
+    return retNode.next;
+};
+```
+
 Kotlin:
 
 ```kotlin

problems/0034.在排序数组中查找元素的第一个和最后一个位置.md

@@ -432,6 +432,7 @@ func getRight(nums []int, target int) int {
 	}
     }
     return border
+
 }
 ```
 

problems/0101.对称二叉树.md

@@ -574,7 +574,87 @@ var isSymmetric = function(root) {
 };
 ```
 
+## Swift:
 
+> 递归
+```swift
+func isSymmetric(_ root: TreeNode?) -> Bool {
+    return _isSymmetric(root?.left, right: root?.right)
+}
+func _isSymmetric(_ left: TreeNode?, right: TreeNode?) -> Bool {
+    // 首先排除空节点情况
+    if left == nil && right == nil {
+        return true
+    } else if left == nil && right != nil {
+        return false
+    } else if left != nil && right == nil {
+        return false
+    } else if left!.val != right!.val {
+        // 进而排除数值不相等的情况
+        return false
+    }
+    
+    // left 和 right 都不为空, 且数值也相等就递归
+    let inSide = _isSymmetric(left!.right, right: right!.left)
+    let outSide = _isSymmetric(left!.left, right: right!.right)
+    return inSide && outSide
+}
+```
+
+> 迭代 - 使用队列
+```swift
+func isSymmetric2(_ root: TreeNode?) -> Bool {
+    guard let root = root else {
+        return true
+    }
+    var queue = [TreeNode?]()
+    queue.append(root.left)
+    queue.append(root.right)
+    while !queue.isEmpty {
+        let left = queue.removeFirst()
+        let right = queue.removeFirst()
+        if left == nil && right == nil {
+            continue
+        }
+        if left == nil || right == nil || left?.val != right?.val {
+            return false
+        }
+        queue.append(left!.left)
+        queue.append(right!.right)
+        queue.append(left!.right)
+        queue.append(right!.left)
+    }
+    return true
+}
+```
+
+> 迭代 - 使用栈
+```swift
+func isSymmetric3(_ root: TreeNode?) -> Bool {
+    guard let root = root else {
+        return true
+    }
+    var stack = [TreeNode?]()
+    stack.append(root.left)
+    stack.append(root.right)
+    while !stack.isEmpty {
+        let left = stack.removeLast()
+        let right = stack.removeLast()
+        
+        if left == nil && right == nil {
+            continue
+        }
+        if left == nil || right == nil || left?.val != right?.val {
+            return false
+        }
+        stack.append(left!.left)
+        stack.append(right!.right)
+        stack.append(left!.right)
+        stack.append(right!.left)
+    }
+    return true
+}
+```
 
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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/0104.二叉树的最大深度.md

@@ -653,5 +653,82 @@ int maxDepth(struct TreeNode* root){
 }
 ```
 
+## Swift
+
+>二叉树最大深度
+```swift
+// 递归 - 后序
+func maxDepth1(_ root: TreeNode?) -> Int {
+    return _maxDepth1(root)
+}
+func _maxDepth1(_ root: TreeNode?) -> Int {
+    if root == nil {
+        return 0
+    }
+    let leftDepth = _maxDepth1(root!.left)
+    let rightDepth = _maxDepth1(root!.right)
+    return 1 + max(leftDepth, rightDepth)
+}
+
+// 层序
+func maxDepth(_ root: TreeNode?) -> Int {
+    guard let root = root else {
+        return 0
+    }
+    var queue = [TreeNode]()
+    queue.append(root)
+    var res: Int = 0
+    while !queue.isEmpty {
+        res += 1
+        for _ in 0 ..< queue.count {
+            let node = queue.removeFirst()
+            if let left = node.left {
+                queue.append(left)
+            }
+            if let right = node.right {
+                queue.append(right)
+            }
+        }
+    }
+    return res
+}
+```
+
+>N叉树最大深度
+```swift
+// 递归
+func maxDepth(_ root: Node?) -> Int {
+    guard let root = root else {
+        return 0
+    }
+    var depth = 0
+    for node in root.children {
+        depth = max(depth, maxDepth(node))
+    }
+    return depth + 1
+}
+
+// 迭代-层序遍历
+func maxDepth1(_ root: Node?) -> Int {
+    guard let root = root else {
+        return 0
+    }
+    var depth = 0
+    var queue = [Node]()
+    queue.append(root)
+    while !queue.isEmpty {
+        let size = queue.count
+        depth += 1
+        for _ in 0 ..< size {
+            let node = queue.removeFirst()
+            for child in node.children {
+                queue.append(child)
+            }
+        }
+    }
+    return depth
+}
+```
+
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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>