ListCycle2

题目描述

Given a linked list, return the node where the cycle begins. If there is no cycle, returnnull.

Follow up:
Can you solve it without using extra space?
算法：

/** * 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,slow=head;        while(fast!=null&&fast.next!=null){            fast=fast.next.next;            slow=slow.next;            if(fast==slow){                slow=head;                while(fast!=slow){                    fast=fast.next;                    slow=slow.next;                }                return fast;            }        }        return null;    }}

分析：
2*(a+b)=a+b+n*(b+c)可得
a=n*(b+c)-b