LeetCode.141_[链表]_环形链表

1.题目

2.解题思路

快慢指针，有环两个指针迟早相遇，没环，快指针走到头就结束

3.代码实现

public class Solution {
    public boolean hasCycle(ListNode head) {
        ListNode first = new ListNode(0);
		first.next = head;
        ListNode quick = first;
        ListNode slow = first;
        while(quick != null && quick.next != null) {
        	quick = quick.next.next;
        	slow = slow.next;
        	if(quick == slow) {
        		return true;
        	}
        }
		return false;
    }
}