题目:
Given a linked list, determine if it has a cycle in it.
思路与步骤:
采用“快慢指针”查检查链表是否含有环。让一个指针一次走一步,另一个一次走两步,如果链表中含有环,快指针会再次和慢指针相遇。
编程实现:
/**
* Definition for singly-linked list.
* class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
public boolean hasCycle(ListNode head) {
if(head==null || head.next==null) return false;
ListNode fast = head;
ListNode slow = head;
while(fast.next!=null && fast.next.next!=null){
fast = fast.next.next;
slow = slow.next;
if(fast==slow) return true;
}
return false;
}
}
注意循环条件!否则会有异常
while(fast.next!=null && fast.next.next!=null)