力扣141. 环形链表

Problem: 141. 环形链表

题目描述

思路

定义快慢指针fast、slow，当fast != null && fast.next != null时fast每次走两步、slow走一步，当fast和slow相遇时，则说明存在环

复杂度

时间复杂度:

$O(n)$

空间复杂度:

$O(1)$

Code

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
class Solution {
    /**
     * Linked List Cycle
     *
     * @param head The head of linked list
     * @return boolean
     */
    public boolean hasCycle(ListNode head) {
        if (head == null || head.next == null) {
            return false;
        }
        ListNode fast = head;
        ListNode slow = head;
        while (fast != null && fast.next != null) {
            fast = fast.next.next;
            slow = slow.next;
            if (fast == slow) {
                return true;
            }
        }
        return false;
    }
}