题目：

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?

思路：

可以参考

LeetCode: 73 Linked List Cycle。

使用快慢两个指针，首先先判断链表是否存在环，如果存在，从相遇点开始，还有另外一个从链表头开始，同时移动一步，由于a = c,再次相遇时，即为环的入口点。否则返回NULL, 不存在环。

复杂度：O(N)

AC Code:

<span style="font-size:14px;">/**
 * 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) {
       if(head == NULL || head->next == NULL) return NULL;
       ListNode* fast = head;
       ListNode* slow = head;

       while(fast != NULL && fast->next != NULL)
       {
           slow = slow->next;
           fast = fast->next->next;
           if(slow == fast)
           {
               slow = head;
               while(slow != fast)
               {
                   slow = slow->next;
                   fast = fast->next;
               }
               return slow;
           }
       }

       return NULL;

    }
};</span>