142. Linked List Cycle II

题目分析

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?

推导过程：
s = x + ny + k
f = x + my + k
f = 2s
(m - 2n)y - x = k
(m-2n)y = k + x
(m-2n)y 正好等于整数倍的环
对应从第 k 个位置走 x 步正好到环的起点

代码

/**
 * 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;
        while(fast != null && slow != null) {
            fast = fast.next;
            slow = slow.next;
            if(fast != null) {
                fast = fast.next;
            }
            if(fast == slow) {
                break;
            }
        }
        if(fast == null) {
            return null;
        }
        fast = head;
        while(fast != slow) {
            fast = fast.next;
            slow = slow.next;
        }
        return fast;
    }
}