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.
Follow up:
Can you solve it without using extra space?
思路：设head距离循环开始点k，循环开始点距离fast和slow第一次相遇点x，slow还要走y到达循环开始点。则有：x+y+k=n; n+x= 2* (k+n); 得到y=k。及相遇点到循环开始点的距离与head到循环开始点的距离相等，那么把slow放到head，fast和slow都用pace=1行走，则在循环开始点两者将相遇。
class Solution { public: ListNode *detectCycle(ListNode *head) { ListNode* slow = head; ListNode* fast = head; bool flag = false; while(fast && fast->next) { slow = slow->next; fast = fast->next->next; if(slow == fast) { slow = head; flag = true; break; } } if(!flag) return NULL; while(fast!=slow){ fast = fast->next; slow = slow->next; } return fast; } };
142. Linked List Cycle II (List; Two-Pointers)
原文：http://www.cnblogs.com/qionglouyuyu/p/4853498.html