链表双指针

难度简单

题目描述

解题思路

快慢指针

Code

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