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.

想法：（1）首先的判斷連結串列中是否有環，若有環進行(2),沒有環就返回NULL

         (2)參考https://www.cnblogs.com/jack204/archive/2011/09/14/2175559.html的分析，

          從連結串列頭到環入口點等於(n-1)迴圈內環+相遇點到環入口點。於是可以從連結串列頭和相遇點分別設一個 指標，每次各走一步，兩個指標必定相遇，且相遇第一點為環入口點。此時返回其中一個指標即可。

/**
 * 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) {
        if(NULL == head || head->next == NULL)
            return NULL;
        
 
bool cycle = false; 
        struct ListNode* fast = head;
        struct ListNode* slow = head;
        while(fast  && fast->next){
            slow = slow->next;
            fast = fast->next->next;
            if(slow == fast){
                
                cycle = true;
                
 
break;
            }
        }
        if(cycle){
            slow = head;
            while(slow != fast){
                slow = slow->next;
                fast = fast->next;
            }
            return slow;
        }

        return NULL;
    }
};