添加（0143.重排链表.md）：增加typescript版本

Author: xiaofei-2020

problems/0143.重排链表.md

@@ -6,6 +6,8 @@
 
 # 143.重排链表
 
+[力扣题目链接](https://leetcode.cn/problems/reorder-list/submissions/)
+
 ![](https://code-thinking-1253855093.file.myqcloud.com/pics/20210726160122.png)
 
 ## 思路
@@ -465,7 +467,81 @@ var reorderList = function(head, s = [], tmp) {
 }
 ```
 
+### TypeScript
+
+> 辅助数组法：
+
+```typescript
+function reorderList(head: ListNode | null): void {
+    if (head === null) return;
+    const helperArr: ListNode[] = [];
+    let curNode: ListNode | null = head;
+    while (curNode !== null) {
+        helperArr.push(curNode);
+        curNode = curNode.next;
+    }
+    let node: ListNode = head;
+    let left: number = 1,
+        right: number = helperArr.length - 1;
+    let count: number = 0;
+    while (left <= right) {
+        if (count % 2 === 0) {
+            node.next = helperArr[right--];
+        } else {
+            node.next = helperArr[left++];
+        }
+        count++;
+        node = node.next;
+    }
+    node.next = null;
+};
+```
+
+> 分割链表法：
+
+```typescript
+function reorderList(head: ListNode | null): void {
+    if (head === null || head.next === null) return;
+    let fastNode: ListNode = head,
+        slowNode: ListNode = head;
+    while (fastNode.next !== null && fastNode.next.next !== null) {
+        slowNode = slowNode.next!;
+        fastNode = fastNode.next.next;
+    }
+    let head1: ListNode | null = head;
+    // 反转后半部分链表
+    let head2: ListNode | null = reverseList(slowNode.next);
+    // 分割链表
+    slowNode.next = null;
+    /**
+    	直接在head1链表上进行插入
+        head1 链表长度一定大于或等于head2,
+        因此在下面的循环中，只要head2不为null, head1 一定不为null
+     */
+    while (head2 !== null) {
+        const tempNode1: ListNode | null = head1!.next,
+            tempNode2: ListNode | null = head2.next;
+        head1!.next = head2;
+        head2.next = tempNode1;
+        head1 = tempNode1;
+        head2 = tempNode2;
+    }
+};
+function reverseList(head: ListNode | null): ListNode | null {
+    let curNode: ListNode | null = head,
+        preNode: ListNode | null = null;
+    while (curNode !== null) {
+        const tempNode: ListNode | null = curNode.next;
+        curNode.next = preNode;
+        preNode = curNode;
+        curNode = tempNode;
+    }
+    return preNode;
+}
+```
+
 ### C
+
 方法三：反转链表
 ```c
 //翻转链表