Given a linked list, return the node where the cycle begins. If there is no cycle, return null.

To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle in the linked list.Note: Do not modify the linked list.Example 1:

Input: head = [3,2,0,-4], pos = 1Output: tail connects to node index 1Explanation: There is a cycle in the linked list, where tail connects to the second node.

Example 2:

Input: head = [1,2], pos = 0Output: tail connects to node index 0Explanation: There is a cycle in the linked list, where tail connects to the first node.

Example 3:

Input: head = [1], pos = -1Output: no cycleExplanation: There is no cycle in the linked list.

Follow up:

Can you solve it without using extra space?

难度：medium

题目：给定一链表，返回环的起始结点。如果无环返回空。为了表示链表中的环，使用整数来表示起如结点位置即尾结点所指向的位置。如果位置为－1则表示无环。

注意：不要修改链表。

思路：快慢指针, 在首次相遇时slow 指针走过的路程为A->B->C, 设A->B为a, B->C为b,C->B为c, fast走过的路程为 A->B->C->B->C->B->C....C。但是快指针的速度是慢指针的两倍，因此快指针走过的路程是慢指针走过路程的两倍，因此 2(a + b) = a + b + c + b + n * (b + c); 所以a % (b + c) = c;

Runtime: 0 ms, faster than 100.00% of Java online submissions for Linked List Cycle II.

Memory Usage: 35.2 MB, less than 100.00% of Java online submissions for Linked List Cycle II.

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