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

To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle in the linked list.

Note: Do not modify the linked list.

 

Example 1:

Input: head = [3,2,0,-4], pos = 1Output: tail connects to node index 1Explanation: There is a cycle in the linked list, where tail connects to the second node.

Example 2:

Input: head = [1,2], pos = 0Output: tail connects to node index 0Explanation: There is a cycle in the linked list, where tail connects to the first node.

Example 3:

Input: head = [1], pos = -1Output: no cycleExplanation: There is no cycle in the linked list.

Follow-up:

Can you solve it without using extra space?

判断环的起始位置，我们还是设置两个不同步长的指针，步长相差为1，经理M步之后相遇，M即为圈的长度。然后我们将两个指针重置为head。其中一个p指针用于定位，另一个q用于遍历，当遍历指针经过M步遍历后回到p，则p为循环开始处。

/** * 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 *p=head;        ListNode *q=head;        int counter=0;        while(p!=NULL&&p->next!=NULL)        {            p=p->next->next;            q=q->next;            counter++;            if(p==q){                p=head;                q=head;                while(1)                {                    for(int i=0;i
     
      next;                    if(p==q) return p;                    else {                        p=p->next;                        q=p;                    }                }            }        }         return NULL;    }};