链表-Java-力扣142. 环形链表 II

1. 题目

题目链接142. 环形链表 II

2. 思路

设置快慢指针，快指针一次走两步，慢指针一次走一步，当他们第一次相遇的时候，将快指针放到头部，然后快指针和慢指针同时一人走一步，然后相遇的节点就是入口位置。

3. 代码

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode detectCycle(ListNode head) {
        ListNode fast = head;
        ListNode slow = head;
        while (fast != null && fast.next != null) {
            fast = fast.next.next;
            slow = slow.next;
            if (fast == slow) {
                break;
            }
        }
        if (fast == null || fast.next == null) return null;
        slow = head;
        while (fast != slow) {
            fast = fast.next;
            slow = slow.next;
        }
        return slow;
    }
}