# \[141]\[简单]\[双指针] 环形链表

## 题目描述

[141. 环形链表](https://leetcode-cn.com/problems/linked-list-cycle/)

给定一个链表，判断链表中是否有环。

为了表示给定链表中的环，我们使用整数 pos 来表示链表尾连接到链表中的位置（索引从 0 开始）。 如果 pos 是 -1，则在该链表中没有环。

示例 1：

```
输入：head = [3,2,0,-4], pos = 1
输出：true
解释：链表中有一个环，其尾部连接到第二个节点。
```

![](/files/-MI7Ns0b_XvuP7lU8E5M)

示例 2：

```
输入：head = [1,2], pos = 0
输出：true
解释：链表中有一个环，其尾部连接到第一个节点。
```

![](/files/-MI7Ns0dH9WiLe8dclyn)

示例 3：

```
输入：head = [1], pos = -1
输出：false
解释：链表中没有环。
```

![](/files/-MI7Ns1FQMk41WOwN1hr)

进阶：

* 你能用 O(1)（即，常量）内存解决此问题吗？

## 解题思路

使用快慢指针, 快指针每次移动两格, 慢指针每次移动一格. 如果没有环, 快指针将会遇到空节点, 从而返回`False`. 如果有环, 在快慢指针都进入环之后, 由于速度的差异, 快慢指针肯定会重合, 重合时返回`True`.

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

class Solution:
    def hasCycle(self, head: ListNode) -> bool:
        slow = fast = head
        while fast is not None and fast.next is not None:
            slow = slow.next
            fast = fast.next.next
            if slow is fast:
                return True
        return False
```


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