# Problem 206: Reverse Linked List

> <https://leetcode.com/problems/reverse-linked-list/>

## 思路

这是一道经典并且基本的题目，理解这道题可以很好地理解 Linked List。先附上一个讲解的视频

> <https://www.youtube.com/watch?v=sYcOK51hl-A>

* 反转链表，主要地是要学会反转指针
* 思路是：1->2->3->4 现在要变成 1<-2<-3<-4，也就是说每个 pointer 都变为指向之前的 node，那么当前的 node 指向一个新设置的 prev node

![](/files/-Lpv9xVHVhgqjE3mmCSS)

![](/files/-Lpv9xVJDgJXSSxoLi8a)

![](/files/-Lpv9xVLMCHmrDN0tLc8)

![](/files/-Lpv9xVNPaXhZS4vP0A5)

## 复杂度

* 时间复杂度：\`O(n)\`
* 空间复杂度：\`O(1)\`

```java
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode reverseList(ListNode head) {
        if (head == null) {
            return head;
        }

        ListNode prev = null;
        while (head != null) {
            ListNode temp = head.next;
            head.next = prev;
            prev = head;
            head = temp;
        }
        return prev;
    }
}
```

## 易错点

1. 始终牢记一点就是，prev和next进行交换
2. 在反转指针的过程中，当断开链接的时候，我们就对剩下的元素失去了track，所以提前用temp存起来，以后用的时候再调出来。
3. head指针其实是current指针，随着他的移动，不停地更新链表
4. 最后结束的时候，其实头是prev指针
5. while循环一遍的时候并没有交换完成一遍，循环好几遍的时候才完成指针的反转


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