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; } } }