63_leetcode_Linked List Cycle II

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?

1:特殊情况；2:判断链表是否有环，如果有环的话，返回true， 并保存fast和slow碰头的节点（meetNode）， 如果没有换直接返回false；3:分别从头节点往后以及meetNode节点向后遍历，当两者相同时，即为cycle开始节点。

    ListNode *detectCycle(ListNode *head)    {        if(head == NULL || head->next == NULL)        {            return NULL;        }                ListNode* meetNode = NULL;        bool flag = hasCycleCore(head, meetNode);                if(flag == false)        {            return NULL;        }        else        {            ListNode *index = head;            while(index != meetNode)            {                index = index->next;                meetNode = meetNode->next;            }            return meetNode;        }    }        bool hasCycleCore(ListNode* head, ListNode* &meetNode)    {        ListNode *slow = head;        ListNode *fast = head;                while(fast && fast->next)        {            slow = slow->next;            fast = fast->next->next;                        if(slow == fast)            {                meetNode = slow;                return true;            }        }                return false;    }