Linked List Cycle II

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

Note: Do not modify the linked list.

Follow up:

Can you solve it without using extra space?

思路：找到一个有环链表的起始节点。定义两个指针，一个指针每次走一步，一个指针每次走两步。这样，当两者相遇的时候说明整个链表含有环。当两者第一次相遇的时候，快指针走的路径是慢指针的两倍。可以得出从第一次相遇的地方到链表环起始节点的距离和从链表head开始到循环起始节点的距离是相等的。这样，再遍历一次链表即可。

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    ListNode *detectCycle(ListNode *head) {        ListNode *slow=head;        ListNode *fast=head;        while(true)        {            if(fast==NULL || fast->next==NULL)                return NULL;            slow=slow->next;            fast=fast->next->next;            if(slow==fast)                break;        }        //跳出循环说明有环        slow=head;        while(slow!=fast)        {            slow=slow->next;            fast=fast->next;        }        return slow;    }};