LeetCode-Linked List Cycle II

作者：disappearedgod

文章出处：http://blog.csdn.net/disappearedgod/article/details/24018961

时间：2014-4-19

题目

Linked List Cycle II

 Total Accepted: 10764 Total Submissions: 35294My Submissions

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

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

 Definition for singly-linked list. public class ListNode {     int val;     ListNode next;     ListNode(int x) {         val = x;         next = null;     } } 

想法：

见《cracking the code interview》P126 中文版
1.一个一步走的slow指针，一个两步走的fast指针
2.如果有环，slow 与fast 一定碰面（如果刚好错过，一定是错过一步（并且是fast(K+1)领先slow(K-1),K为环中从相交点开始计算的数字），但是，他们的上一步（均为k-1）正好相遇）
3.假设Fast走了2N步，那么slow走了N步。
如果环一共有M个节点，当

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

返回

LeetCode Solution(持续更新，java>c++)