快指针速度2 , 慢指针速度1
相对速度1,有环必然相遇
public class Solution {
public boolean hasCycle(ListNode head) {
ListNode fast = head,slow = head;
while(fast!=null && fast.next!=null){
slow = slow.next;
fast = fast.next.next;
if(slow == fast){
return true;
}
}
return false;
}
}
[LeetCode] 141. Linked List Cycle 单链表判圆算法
原文:https://www.cnblogs.com/Poceer/p/10954550.html