leetcode Linked List Cycle题解

最新推荐文章于 2025-05-07 22:15:24 发布

leo_weile

最新推荐文章于 2025-05-07 22:15:24 发布

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本文介绍了一种使用快慢指针方法来判断链表是否存在环的有效算法。通过一个指针每次移动一步，另一个指针每次移动两步，可以高效地检测出链表中是否存在环，并附带Java代码实现。

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中文描述：

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.

Follow up:

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

中文理解：给出一个链表，求出该链表是否存在环。

解题思路：使用快慢指针方法，一个指针一次走一步，一个指针一次走两步，若两个指针值在某一处相同了，那该链表存在环，若任意一个指针走到了null，则该链表不存在环。

代码（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) {
        if(head==null || head.next==null || head.next.next==null)return false;
        else{
            ListNode pFast=head.next.next;
            ListNode pSlow=head.next;
            while(pFast!=pSlow){
                if(pFast.next==null || pFast.next.next==null)return false;
                pFast=pFast.next.next;
                if(pSlow.next==null)return false;
                pSlow=pSlow.next;
            }
            return true;
        }
    }
}