题目描述

• 相对于环形链表，这里要求找到环的起点
• 难点在于 O(1)，否则可以直接哈希表冲
  

思路 & 代码

• 找出快慢指针的路程关系，得出结论（详见代码注释）

/*** Definition for singly-linked list.* class ListNode {*     int val;*     ListNode next;*     ListNode(int x) {*         val = x;*         next = null;*     }* }*/// 相遇时结束：fastDistance = 2 * slowDistance（两倍路程
// fastDistance = slowDistance + n * b （比 slow 多走 n 圈）
// =》slowDistance = n * b
public class Solution {public ListNode detectCycle(ListNode head){// 空链表ListNode slow = head, fast = head;while(true){// 无环情况if(fast == null || fast.next == null){return null;}slow = slow.next;fast = fast.next.next;if(fast == slow){break;}}// 令 fastDistance = 0，此时 slowDistance = n * b// 只要令 slowDistance = a + n * b，即可走到环起点fast = head;while(fast != slow){fast = fast.next;slow = slow.next;}return fast;}
}

更新版

• 虽然但是，感觉有稍微优化了一下 while 结构= =

public class Solution {public ListNode detectCycle(ListNode head) {ListNode slow = head, fast = head;do {if(fast == null || fast.next == null) {return null;}slow = slow.next;fast = fast.next.next;} while(slow != fast);fast = head;while(fast != slow) {fast = fast.next;slow = slow.next;}return fast;}
}

三刷 - 再更新

• 选取了一个我更喜欢的结构！
• 成环情况，直接开找，然后return

public 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(fast == slow) {fast = head;while(fast != slow) {fast = fast.next;slow = slow.next;}return fast;}}return null;}
}