Linked List Cycle

Given a linked list, determine if it has a cycle in it.

Follow up:

Can you solve it without using extra space?

题意：判断链表中是否有环

思路：采用快慢指针的方法，快指针走两步，慢指针走一步，如果有环，则总有相遇的时候即快慢指针相等的时候

/** * 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) {        ListNode fast = head;     ListNode slow = head;               while (fast != null && fast.next != null) {        fast = fast.next.next;slow = slow.next;        if (fast == slow) {return true;}}       return false;    }   }

输入链表为空时，没有环，返回false。
因为fast指针比slow指针走得快，所以只要判断fast指针是否为空就行。由于fast指针一次走两步，所以当fast.next可能已经为空（当fast为尾结点时），fast.next.next将会导致NullPointerException异常，所以在while循环中要判断fast.next是否为空