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60 lines
1.5 KiB
Markdown
60 lines
1.5 KiB
Markdown
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# LeetCode 第 206 号问题：反转链表
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> 本文首发于公众号「五分钟学算法」，是[图解 LeetCode ](<https://github.com/MisterBooo/LeetCodeAnimation>)系列文章之一。
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>
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> 个人网站：[https://www.cxyxiaowu.com](https://www.cxyxiaowu.com)
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题目来源于 LeetCode 上第 206 号问题：反转链表。题目难度为 Easy，目前通过率为 45.8% 。
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### 题目描述
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反转一个单链表。
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**示例:**
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```
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输入: 1->2->3->4->5->NULL
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输出: 5->4->3->2->1->NULL
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```
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**进阶:**
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你可以迭代或递归地反转链表。你能否用两种方法解决这道题？
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### 题目解析
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设置三个节点`pre`、`cur`、`next`
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- （1）每次查看`cur`节点是否为`NULL`，如果是，则结束循环，获得结果
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- （2）如果`cur`节点不是为`NULL`，则先设置临时变量`next`为`cur`的下一个节点
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- （3）让`cur`的下一个节点变成指向`pre`，而后`pre`移动`cur`，`cur`移动到`next`
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- （4）重复（1）（2）（3）
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### 动画描述
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### 代码实现
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```
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class Solution {
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public:
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ListNode* reverseList(ListNode* head) {
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ListNode* pre = NULL;
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ListNode* cur = head;
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while(cur != NULL){
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ListNode* next = cur->next;
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cur->next = pre;
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pre = cur;
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cur = next;
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}
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return pre;
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}
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};
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```
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