腾讯精选50题（3）--141. 环形链表

解法1：双指针

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
/*双指针：一快一满，
         slow每次只走一步；fast每次走两步；
如果是有循环的环，快的总会追上慢的，否则则没有*/
public class Solution {
    public boolean hasCycle(ListNode head) {
        if (head == null||head.next==null) return false;
        ListNode slow = head;
        ListNode fast = head.next;
        while(slow!=fast ){
            if(fast==null||fast.next==null){
                return false;
            }
                slow = slow.next;
                fast = fast.next.next;  
        }
        return true;
    }
}

解法2：Hash table