From 23b9557407ed8b09a38844c873598b5375eac18e Mon Sep 17 00:00:00 2001
From: Omooo <869759698@qq.com>
Date: Mon, 1 Jun 2020 07:54:26 +0800
Subject: [PATCH] =?UTF-8?q?Update=20=E9=93=BE=E8=A1=A8=E7=9B=B8=E5=85=B3.m?=
 =?UTF-8?q?d?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 blogs/Algorithm/剑指 Offer/链表相关.md | 50 ++++++++++++++++++--
 1 file changed, 47 insertions(+), 3 deletions(-)

diff --git a/blogs/Algorithm/剑指 Offer/链表相关.md b/blogs/Algorithm/剑指 Offer/链表相关.md
index 6c194dd..5b6c521 100644
--- a/blogs/Algorithm/剑指 Offer/链表相关.md	
+++ b/blogs/Algorithm/剑指 Offer/链表相关.md	
@@ -2,11 +2,11 @@
 链表相关
 ---
 
-[06: 从头到尾打印链表](https://leetcode-cn.com/problems/cong-wei-dao-tou-da-yin-lian-biao-lcof/)
+[06. 从头到尾打印链表](https://leetcode-cn.com/problems/cong-wei-dao-tou-da-yin-lian-biao-lcof/)
 
 ```java
 class Solution {
-    public int[] reversePrint(ListNode head) {
+    public int[s] reversePrint(ListNode head) {
         ListNode temp = head;
         int size = 0;
         while (temp != null) {
@@ -43,7 +43,7 @@ class Solution {
 }
 ```
 
-[22:链表中倒数第 k 个节点](https://leetcode-cn.com/problems/lian-biao-zhong-dao-shu-di-kge-jie-dian-lcof/)
+[22. 链表中倒数第 k 个节点](https://leetcode-cn.com/problems/lian-biao-zhong-dao-shu-di-kge-jie-dian-lcof/)
 
 ```java
 class Solution {
@@ -78,3 +78,47 @@ class Solution {
 }
 ```
 
+[24. 反转链表](https://leetcode-cn.com/problems/fan-zhuan-lian-biao-lcof/)
+
+```java
+    class Solution {
+        public ListNode reverseList(ListNode head) {
+            if (head == null || head.next == null) {
+                return head;
+            }
+            ListNode h1 = head;
+            ListNode h2 = head.next;
+            ListNode h3 = null;
+            h1.next = null;
+            while (h2 != null) {
+                h3 = h2.next;
+                h2.next = h1;
+                h1 = h2;
+                h2 = h3;
+            }
+            return h1;
+        }
+    }
+```
+
+```java
+    class Solution {
+        public ListNode reverseList(ListNode head) {
+            // 递归终止条件是当前为空，或者下一个节点为空
+            if (head == null || head.next == null) {
+                return head;
+            }
+            // 这里的 h1 就是最后一个节点
+            ListNode h1 = reverseList(head.next);
+            // 如果链表是 1->2->3->4->5，那么此时的 cur 就是 5
+            // 而 head 是4，head的 下一个是 5，下下一个是空
+            // 所以 head.next.next 就是 5->4
+            head.next.next = head;
+            // 防止链表循环，需要将 head.next 设置为空
+            head.next = null;
+            // 每层递归函数都返回 h1，也就是最后一个节点
+            return h1;
+        }
+    }
+```
+