141. Linked List Cycle

最新推荐文章于 2025-05-07 22:36:04 发布

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最新推荐文章于 2025-05-07 22:36:04 发布

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本文链接：https://blog.youkuaiyun.com/zzzyqsama/article/details/103023389

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本文介绍了一种使用快慢指针的方法来判断链表中是否存在循环。通过两个速度不同的指针遍历链表，如果存在循环，快指针最终会追上慢指针；反之则返回False。此方法在链表节点数未知的情况下尤其有效。

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Linked List Cycle
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.
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:
        fast = low = head
        while fast and fast.next:
            fast = fast.next.next
            low = low.next 
            if low == fast:
                return True
        return False

java:

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public boolean hasCycle(ListNode head) {
        ListNode fast = head;
        ListNode low = head;
        while(fast!=null && fast.next!=null){
            fast= fast.next.next;
            low = low.next;
            if(low == fast)
                return true;
        }
        return false;
    }
}