# 141. Linked List Cycle 环形链表

@TOC

[LeetCode]

Total Accepted: 102417 Total Submissions: 277130 Difficulty: Easy

## # 题目描述

Given a linked list, determine if it has a cycle in it.

To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle in the linked list.

Example 1:

Input: head = [3,2,0,-4], pos = 1 Output: true Explanation: There is a cycle in the linked list, where tail connects to the second node.

Example 2:

``````Input: head = [1,2], pos = 0
Output: true
Explanation: There is a cycle in the linked list, where tail connects to the first node.
``````

Example 3:

``````Input: head = [1], pos = -1
Output: false
Explanation: There is no cycle in the linked list.
``````

Can you solve it using O(1) (i.e. constant) memory?

## # 解题方法

### # 双指针

Java代码如下：

``````public class Solution {
while(slow!=null){
if(fast.next==null || fast.next.next==null) return false;
fast=fast.next.next;
slow=slow.next;
if(fast==slow) break;
}
return true;
}
}
``````

AC:1ms

``````	public class Solution {
while(slow!=fast){
if(fast.next==null || fast.next.next==null) return false;
fast=fast.next.next;
slow=slow.next;
}
return true;
}
}
``````

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

class Solution(object):
"""
:rtype: bool
"""
while fast and fast.next:
fast = fast.next.next
slow = slow.next
if slow == fast:
return True
return False
``````

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

class Solution(object):
"""
:rtype: bool
"""
while fast and fast.next:
if fast == slow:
return True
fast = fast.next.next
slow = slow.next
return False
``````

``````/**
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
while (fast && fast->next) {
fast = fast->next->next;
slow = slow->next;
if (fast == slow)
return true;
}
return false;
}
};
``````

### # 保存已经走过的路径

``````public class Solution {
HashSet<ListNode> hash=new HashSet();
return true;
}else{