diff --git a/problems/0143.重排链表.md b/problems/0143.重排链表.md
index 2b4e68b7..07c3bb40 100644
--- a/problems/0143.重排链表.md
+++ b/problems/0143.重排链表.md
@@ -351,6 +351,110 @@ class Solution:
 Go：
 
 JavaScript：
+```javascript
+// 方法一 使用数组存储节点
+var reorderList = function(head, s = [], tmp) {
+    let cur = head;
+    // list是数组，可以使用下标随机访问
+    const list = [];
+    while(cur != null){
+        list.push(cur);
+        cur = cur.next;
+    }
+    cur = head; // 重新回到头部
+    let l = 1, r = list.length - 1; // 注意左边是从1开始
+    let count = 0;
+    while(l <= r){
+        if(count % 2 == 0){
+            // even
+            cur.next = list[r];
+            r--;
+        } else {
+            // odd
+            cur.next = list[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.length % 2 == 0){
+        cur.next = list[l];
+        cur = cur.next;
+    }
+    // 注意结尾要结束一波
+    cur.next = null;
+}
+
+// 方法二 使用双端队列的方法来解决 js中运行很慢
+var reorderList = function(head, s = [], tmp) {
+    // js数组作为双端队列
+    const deque = [];
+    // 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
+    let cur = head.next;
+    while(cur != null){
+        deque.push(cur);
+        cur = cur.next;
+    }
+    cur = head;  // 回到头部
+    let count = 0;
+    while(deque.length !== 0){
+        if(count % 2 == 0){
+            // even，取出队列右边尾部的值
+            cur.next = deque.pop();
+        } else {
+            // odd, 取出队列左边头部的值
+            cur.next = deque.shift();
+        }
+        cur = cur.next;
+        count++;
+    }
+    cur.next = null;
+}
+
+//方法三 将链表分割成两个链表，然后把第二个链表反转，之后在通过两个链表拼接成新的链表
+var reorderList = function(head, s = [], tmp) {
+    const reverseList = head => {
+        let headNode = new ListNode(0);
+        let cur = head;
+        let next = null;
+        while(cur != null){
+            next = cur.next;
+            cur.next = headNode.next;
+            headNode.next = cur;
+            cur = next;
+        }
+        return headNode.next;
+    }
+
+    let fast = head, slow = head;
+    //求出中点
+    while(fast.next != null && fast.next.next != null){
+        slow = slow.next;
+        fast = fast.next.next;
+    }
+    //right就是右半部分 12345 就是45  1234 就是34
+    let right = slow.next;
+    //断开左部分和右部分
+    slow.next = null;
+    //反转右部分 right就是反转后右部分的起点
+    right = reverseList(right);
+    //左部分的起点
+    let left = head;
+    //进行左右部分来回连接 
+    //这里左部分的节点个数一定大于等于右部分的节点个数 因此只判断right即可
+    while (right != null) {
+        let curLeft = left.next;
+        left.next = right;
+        left = curLeft;
+
+        let curRight = right.next;
+        right.next = left;
+        right = curRight;
+    }
+}
+```
 
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