solution/0200-0299/0206.Reverse Linked List/README.md · buelev/leetcode

# [206. 反转链表](https://leetcode-cn.com/problems/reverse-linked-list)

[English Version](/solution/0200-0299/0206.Reverse%20Linked%20List/README_EN.md)

## 题目描述

反转一个单链表。

示例:

进阶: 
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？

## 解法

定义指针 `p`、`q` 分别指向头节点和下一个节点，`pre` 指向头节点的前一个节点。

遍历链表，改变指针 `p` 指向的节点的指向，将其指向 `pre` 指针指向的节点，即 `p.next = pre`。然后 `pre` 指针指向 `p`，`p`、`q` 指针往前走。

当遍历结束后，返回 `pre` 指针即可。

### **Python3**

```python
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def reverseList(self, head: ListNode) -> ListNode:
        pre, p = None, head
        while p:
            q = p.next
            p.next = pre
            pre = p
            p = q
        return pre
```

### **Java**

迭代版本：

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

递归版本：

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

### **JavaScript**

```js
/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} head
 * @return {ListNode}
 */
var reverseList = function (head) {
    let pre = null;
    for (let p = head; p; ) {
        let q = p.next;
        p.next = pre;
        pre = p;
        p = q;
    }
    return pre;
};
```

### **Go**

```go
/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func reverseList(head *ListNode) *ListNode {
	var pre *ListNode
	for p := head; p != nil; {
		q := p.Next
		p.Next = pre
		pre = p
		p = q
	}
	return pre
}
```

### **C++**

```cpp
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        ListNode* pre = nullptr;
        ListNode* p = head;
        while (p)
        {
            ListNode* q = p->next;
            p->next = pre;
            pre = p;
            p = q;
        }
        return pre;
    }
};
```

### **C#**

```cs
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int val=0, ListNode next=null) {
 *         this.val = val;
 *         this.next = next;
 *     }
 * }
 */
public class Solution {
    public ListNode ReverseList(ListNode head) {
        ListNode pre = null;
        for (ListNode p = head; p != null;)
        {
            ListNode t = p.next;
            p.next = pre;
            pre = p;
            p = t;
        }
        return pre;
    }
}
```

### **...**

```

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