141. Linked List Cycle

最新推荐文章于 2025-08-08 22:01:10 发布

转载最新推荐文章于 2025-08-08 22:01:10 发布 · 56 阅读

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原文链接：http://www.cnblogs.com/yrbbest/p/4438817.html

文章标签：

#数据结构与算法 #java

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，并详细解释了算法的时间复杂度和空间复杂度为O(n)和O(1)，并提供了Java代码实现。

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题目:

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

Follow up:
Can you solve it without using extra space?

链接： http://leetcode.com/problems/linked-list-cycle/

题解：

链表检测是否有环，使用快慢指针。

Time Complexity - O(n)， Space Complexity - O(1)。

public class Solution {
    public boolean hasCycle(ListNode head) {
        if(head == null || head.next == null)
            return false;
        ListNode slow = head, fast = head;            
        
        while(fast != null && fast.next != null){
            fast = fast.next.next;
            slow = slow.next;
            if(slow == fast)
                return true;
        }
        
        return false;
    }
}

Update:

/**
 * 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)
            return false;
        ListNode fast = head, slow = head;
        
        while(fast != null && fast.next != null) {
            fast = fast.next.next;
            slow = slow.next;
            
            if(slow == fast)
                return true;
        }
        
        return false;
    }
}

二刷:

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) return false;
        ListNode fast = head, slow = head;
        
        while (fast != null && fast.next != null) {
            fast = fast.next.next;
            slow = slow.next;
            if (fast == slow) return true;
        }
        return false;
    }
}