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.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
struct ListNode
{
	int val;
	ListNode *next;
	ListNode(int x) : val(x), next(NULL) {}
};

//这题是Linked list Cycle的变形题，首先用fast,slow指针，判断是否是Cycle list,
//然后，再将slow，指向head头结点，然后继续遍历，当fast == slow的时候，那么，slow的值就是Cycle的入口点。
class Solution {
public:
    ListNode *detectCycle(ListNode *head) {
        if(head == NULL || head->next == NULL) return head;
		ListNode *fast = head;
		ListNode *slow = head;
		bool isCycle = false;
		while(fast != NULL && slow != NULL)
		{
			slow = slow->next;
			if(fast->next == NULL) return NULL;
			fast = fast->next->next;
			if(fast == slow)
			{
				isCycle = true;
				break;
			}
		}
		if(!isCycle) return NULL;
		slow = head;
		while(fast != slow)
		{
			fast = fast->next;
			slow = slow->next;
		}
		return slow;
    }
};

leetcode - Linked List Cycle II