142. Linked List Cycle II (List; Two-Pointers)

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;
    }
};