From 8f1fe3c1b747716d98f22e4f14b2e25e94a108d1 Mon Sep 17 00:00:00 2001
From: Arthur <gpan20020205@gmail.com>
Date: Thu, 21 Oct 2021 15:10:36 +0100
Subject: [PATCH] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200206.=E7=BF=BB=E8=BD=AC?=
 =?UTF-8?q?=E9=93=BE=E8=A1=A8.md=20C=E8=AF=AD=E8=A8=80=E7=89=88=E6=9C=AC?=
MIME-Version: 1.0
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---
 problems/0206.翻转链表.md | 39 +++++++++++++++++++++++++++++++++++
 1 file changed, 39 insertions(+)

diff --git a/problems/0206.翻转链表.md b/problems/0206.翻转链表.md
index 45196386..08495259 100644
--- a/problems/0206.翻转链表.md
+++ b/problems/0206.翻转链表.md
@@ -371,6 +371,45 @@ func reverse(pre: ListNode?, cur: ListNode?) -> ListNode? {
 }
 ```
 
+C:
+双指针法：
+```c
+struct ListNode* reverseList(struct ListNode* head){
+    //保存cur的下一个结点
+    struct ListNode* temp;
+    //pre指针指向前一个当前结点的前一个结点
+    struct ListNode* pre = NULL;
+    //用head代替cur，也可以再定义一个cur结点指向head。
+    while(head) {
+        //保存下一个结点的位置
+        temp = head->next;
+        //翻转操作
+        head->next = pre;
+        //更新结点
+        pre = head;
+        head = temp;
+    }
+    return pre;
+}
+```
+
+递归法：
+```c
+struct ListNode* reverse(struct ListNode* pre, struct ListNode* cur) {
+    if(!cur)
+        return pre;
+    struct ListNode* temp = cur->next;
+    cur->next = pre;
+    //将cur作为pre传入下一层
+    //将temp作为cur传入下一层，改变其指针指向当前cur
+    return reverse(cur, temp);
+}
+
+struct ListNode* reverseList(struct ListNode* head){
+    return reverse(NULL, head);
+}
+```
+
 -----------------------
 * 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
 * B站视频：[代码随想录](https://space.bilibili.com/525438321)