142. Linked List Cycle II(重排链表)

题目链接：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.

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 = 1
Output: tail connects to node index 1
Explanation: There is a cycle in the linked list, where tail connects to the second node.

Example 2:

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

Example 3:

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

Follow up:
Can you solve it without using extra space?

思路：快慢指针

具体的严密证明我没有去做，我只是找几个带环的例子进行了实际的测试，发现，当fast

和slow都从head出发，fast以2倍速，slow以一倍速，当两个节点相遇，然后fast再改为一倍速，

slow仍是一倍速，那么下次fast和slow相遇的节点就是 环的入口节点。

AC 0ms Java：

/**
 * 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(head==null||head.next==null)
            return null;
        ListNode fast=head;
        ListNode slow=head;
        boolean isCircle=false;
        while(fast!=null&&fast.next!=null){
            fast=fast.next.next;
            slow=slow.next;
            if(fast==slow){
                isCircle=true;
                break;
            }
        }
        if(!isCircle)
            return null;
        fast=head;
        while(isCircle&&fast!=slow){
            fast=fast.next;
            slow=slow.next;
        }
        return fast;
    }
}