Linked List Cycle II问题及解法

问题描述：

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?

问题分析：

使用双指针fast，slow，判断是否有环；若有环，则设置链表起点到环起点的距离为s，环起点到fast和slow相遇点的距离为m，环的周长为r。则可列出如下等式：s = r - m + nr(n >=0,指slow从环起点开始共走了n圈)。有了等式，求解就不成问题了。

过程详见代码：

/** * 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 * fast = head, *slow = head;while (fast!= NULL && fast->next != NULL){fast = fast->next->next;slow = slow->next;if (slow == fast){while (slow != head){head = head->next;slow = slow->next;}return head;}}return NULL;    }};