> For the complete documentation index, see [llms.txt](https://wentao-shao.gitbook.io/leetcode/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://wentao-shao.gitbook.io/leetcode/linked-list/206.reverse-linked-list.md).

# 206.Reverse-Linked-List

## 206. Reverse Linked List

## 题目地址

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

<https://www.jiuzhang.com/solutions/palindrome-partitioning-ii/>

## 题目描述

```
Reverse a linked list.
```

## 代码

### Approach 1: 使用while Iterative

1. 保存next 节点
2. curr的next节点指向prev节点
3. prev保存curr节点
4. curr保存next节点

```java
public class Solution {
    public ListNode reverseList(ListNode head) {
    ListNode prev = null;
    ListNode curr = head;

    // prev => cur => next
    while(curr != null) {
      ListNode next = curr.next;
      curr.next = prev;
      prev = curr;
      curr = next;
    }

    head = prev;

    return head;
  }
}
```

### Approach #2 Recursive

n1 → … → nk-1 → **nk** → nk+1 ← … ← nm

**Complexity analysis**

* Time complexity : &#x4F;*(\_n). Assume that n is the list's length, the time complexity is O*(*n*).
* Space complexity : &#x4F;*(\_n*). The extra space comes from implicit stack space due to recursion. The recursion could go up to n\_ levels deep.

```java
public ListNode reverseList(ListNode head) {
  if (head == null || head.next == null) return head;
  //  head => next <= next.next <=
  ListNode p = reverseList(head.next);
  head.next.next = head;
  head.next = null;
  return p;
}
```