Leetcode.142-Linked-list-cycle-ii(环形链表II)

环形链表II

思路 https://www.cnblogs.com/springfor/p/3862125.html

https://blog.csdn.net/u010292561/article/details/80444057

假设周长为 S

AB + BC + n*S = 2 * ( AB + BC )

=> AB = BC + n*S

只要知道了 C 点， 就可以知道 B点了。C点通过快慢指针获得，然后 让一个指针在A点出发，另外一个在C点出发，经过 n 圈，相遇的点就是 AB 点。

  /**
     * 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) {

            if (null == head || null == head.next) {
                return null;
            }

            ListNode p1 = head;
            ListNode p2 = head;
            boolean sign = false;
            while (null != p2 && null != p2.next) {
                p1 = p1.next;
                p2 = p2.next.next;
                if (p1 == p2) {
                    sign = true;
                    break;
                }
            }

            p1 = head;
            if (sign) {
                while (p1 != p2) {
                    p1 = p1.next;
                    p2 = p2.next;
                }
                return p1;
            } else {
                return null;
            }
        }
    }