问题描述：
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?
代码实现：
/**
* 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) return null;
ListNode fast=head;
ListNode slow=head;
do{
if(fast!=null) fast=fast.next;
if(fast!=null){
fast=fast.next;
}else{
return null;
}
slow=slow.next;
}while(fast != slow);
slow=head;
while(fast!=slow){
fast=fast.next;
slow=slow.next;
}
return slow;
}
}
原创文章，作者：Maggie-Hunter，如若转载，请注明出处：https://blog.ytso.com/tech/aiops/7202.html