Problem: 142. 环形链表 II

思路

👨‍🏫 参考题解

Code

⏰ 时间复杂度: $O(n)$
🌎 空间复杂度: $O(1)$

/**
/*** 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){ListNode f = head;ListNode s = head;while (f != null && f.next != null){f = f.next.next;s = s.next;if (f == s)// 到相遇点了{while (s != head)//两者相等即走到了入环点{s = s.next;// s 走 相遇点到入环点 的路head = head.next;// head 走 起点到入环点的路}return s;}}return null;}
}