Leetcode 142. Linked List Cycle IIJAVA语言

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.

题意：不破坏原链表的情况下判断有没有环，，，，，，

/*** 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) {/不让修改原来的结构。本来还想用断链法.if(head==null || head.next==null ||head.next.next==null)return null;ListNode fast=head.next.next;ListNode slow=head.next;///判断有没有环。找到相遇点while(fast!=slow){if(fast.next!=null && fast.next.next!=null){fast=fast.next.next;slow=slow.next;}else{return null;}}fast=head;while(fast!=slow){fast=fast.next;slow=slow.next;}return fast;}
}

PS：依然是快慢指针。先让fast走两步和slow走一步，判断有没有环先，有环的话就会相遇。当他们相遇后，让fast回到头，此时走一步。直到和slow相遇。则此时的相遇点位环入口！！！！！详情见https://www.nowcoder.com/questionTerminal/253d2c59ec3e4bc68da16833f79a38e4

里面有一段推导

转载于:https://blog.51cto.com/fulin0532/1905477

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