标签：nod ext turn lis ace style fast odi next
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?
//之间做过 有推倒
public class Solution {
public ListNode detectCycle(ListNode head) {
ListNode slow = head;
ListNode fast = head;
while (fast!=null && fast.next!=null){
fast = fast.next.next;
slow = slow.next;
if (fast == slow){
ListNode slow2 = head;
while (slow2 != slow){
slow = slow.next;
slow2 = slow2.next;
}
return slow;
}
}
return null;
}
}
标签：nod ext turn lis ace style fast odi next
原文地址：https://www.cnblogs.com/MarkLeeBYR/p/10677650.html
