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?

这题是在　Linked List Cycle II 的基础上(http://blog.csdn.net/ayst123/article/details/38774927)，把return true换成return findnode()即可

假设fast和slow相遇在环中A点，如果slow 走了n 步，则fast走了2n步。此时，如果fast不动，slow再走n步就会追上n步。再如果fast从head开始，则fast和slow还会在A点相遇。　很容易想到，他们之前肯定还会再遇到至少一次，这个点就是环入口。

C++ 

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    ListNode* findnode(ListNode* head, ListNode* slow){        ListNode* fast = head;        while (fast!=slow){            fast = fast->next;            slow = slow->next;        }        return fast;    }        ListNode *detectCycle(ListNode *head) {        ListNode* fast = head;        ListNode* slow = head;        while (fast!=NULL and fast->next!=NULL){            fast = fast->next->next;            slow = slow->next;            if (fast==slow){                return findnode(head,slow);            }        }        return NULL;    }};