Linked-list-cycle

原创于 2018-06-19 18:32:40 发布 · 160 阅读

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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。通过一个快速节点和一个慢速节点遍历链表，如果两者相遇则表明链表有循环。

题目描述

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

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

code:
/**
 * 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)
            return false;
        ListNode snode = head;
        ListNode fnode = head;
        while(fnode != null && fnode.next != null)
        {
            fnode = fnode.next.next;
            snode = snode.next;
            if(snode == fnode)
                return true;
        }
        return false;
    }
}

use a fast node and a slow node, if the fast node can match the slow node, then it can be concluded that it has a cycle.