题目：给定一个单链表，推断链表是否存在环路（是否能不使用额外内存空间）

算法：快慢指针

原理：每次，快指针走一步，慢指针走两步，若链表存在循环。则快慢指针终于必然会在某个节点汇合。否则直到遍历完整个链表都不会汇合

/** * 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 (null==head || null==head.next) {	    		return false;	    	}	    		    	ListNode fast = head;  // fast pointer take one step	    	ListNode slow = head.next.next;  // slow pointer take two steps	    	while (null!=fast && null!=slow) {	    		if (fast == slow) {	    			// fast pointer meet slow pointer, it means list has cycle!	    			return true;	    		}	    			    		if (null == slow.next) {	    			// it mean slow pointer has reach the list tail! No cycle!	    			return false;	    		}	    		else {	    			fast = fast.next;		    		slow = slow.next.next;	    		}	    	}	    	return false;	    }}