From cd7ef344e76cfb913256e65076de5d62cf47080c Mon Sep 17 00:00:00 2001
From: bourne-3 <595962708@qq.com>
Date: Mon, 6 Sep 2021 22:36:27 +0800
Subject: [PATCH] =?UTF-8?q?=E6=B7=BB=E5=8A=A0java=E5=AE=9E=E7=8E=B0143?=
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---
 problems/0143.重排链表.md | 30 ++++++++++++++++++++++++++++++
 1 file changed, 30 insertions(+)

diff --git a/problems/0143.重排链表.md b/problems/0143.重排链表.md
index 62d4cb3f..2b4e68b7 100644
--- a/problems/0143.重排链表.md
+++ b/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;
@@ -259,6 +260,35 @@ public class ReorderList {
         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;
+    }
+}
 
 ```