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?

解题思路

设链表长度为n，头结点与循环节点之间的长度为k。定义两个指针slow和fast，slow每次走一步，fast每次走两步。当两个指针相遇时，有：

• fast = slow * 2
• fast - slow = (n - k)的倍数
  由上述两个式子能够得到slow为（n-k）的倍数

两个指针相遇后，slow指针回到头结点的位置，fast指针保持在相遇的节点。此时它们距离循环节点的距离都为k，然后以步长为1遍历链表，再次相遇点即为循环节点的位置。

实现代码

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */

 //Runtime:16 ms
class Solution {
public:
    ListNode *detectCycle(ListNode *head) {
        if (head == NULL)
        {
            return NULL;
        }

        ListNode *slow = head;
        ListNode *fast = head;
        while (fast->next && fast->next->next)
        {
            slow = slow->next;
            fast = fast->next->next;
            if (fast == slow)
            {
                break;
            }
        }

        if (fast->next && fast->next->next)
        {
            slow = head;
            while (slow != fast)
            {
                slow = slow->next;
                fast = fast->next;
            }

            return slow;
        }

        return NULL;
    }
};