Linked List Cycle II

题目来自：https://leetcode.com/problems/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?

代码

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

解释：

假设

我们假设环的开始地方是$x$而环的长度为$y$。快指针和慢指针相遇在离远点为$t$的地方
那么有: