Given a linked list, determine if it has a cycle in it.
Follow up:
Can you solve it without using extra space?
题意：
判断一个链表是否有环。
解决方案：
双指针，快指针每次走两步，慢指针每次走一步，
如果有环，快指针和慢指针会在环内相遇，fast == slow，这时候返回true。
如果没有环，返回false.
/**
* 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, slow = head;
if(head == null || head.next == null) return false; while(fast != null && fast.next != null){
fast = fast.next.next;
slow = slow.next; if(fast == slow){
return true;
}
} return false;
}
}