From 7b1646ab62451c944b308c081e018ab7f7c9111e Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E5=94=90=E6=A0=91=E6=A3=AE?= <14021051@buaa.edu.cn>
Date: Sat, 22 Sep 2018 19:27:37 +0800
Subject: [PATCH] Create 206. Reverse Linked List.md

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+# [206. Reverse Linked List](https://leetcode.com/problems/reverse-linked-list/description/)
+# 思路
+## 思路一：迭代
+先设置一个头结点List_head，其next指向NULL。然后从待翻转链表中一次取一个节点p出来，将p的next指向List_head的next，List_head的next指向p。
+循环上述操作直到p为NULL。   
+## 思路二：递归
+也可以采用递归的方式：
+* 递归出口：若head == NULL 或者 head -> next == NULL，直接返回head即可；
+* 递归主体：用q记录head的下一个节点，然后令p等于reverseList(q)，则p的最后一个非空节点就是q，将q的next令为head，head的next令为NULL，再返回p即可。
+
+# C++
+## 思路一
+```
+/**
+ * Definition for singly-linked list.
+ * struct ListNode {
+ *     int val;
+ *     ListNode *next;
+ *     ListNode(int x) : val(x), next(NULL) {}
+ * };
+ */
+class Solution {
+public:
+    ListNode* reverseList(ListNode* head) {
+        ListNode *List_head = new ListNode(0);
+        List_head -> next = NULL;
+        
+        ListNode *p = head, *tmp;
+        while(p){
+            tmp = p -> next;
+            p -> next = List_head -> next;
+            List_head -> next = p;
+            p = tmp;
+        }
+        return List_head -> next;
+        
+    }
+};
+```
+## 思路二
+```
+/**
+ * Definition for singly-linked list.
+ * struct ListNode {
+ *     int val;
+ *     ListNode *next;
+ *     ListNode(int x) : val(x), next(NULL) {}
+ * };
+ */
+class Solution {
+public:
+    ListNode* reverseList(ListNode* head) {
+        if(head == NULL || head -> next == NULL) return head;
+        ListNode *p, *q;
+        
+        q = head -> next;
+        p = reverseList(q);
+        q -> next = head;
+        head -> next = NULL;
+        return p;
+    }
+};
+```