Given a linked list, determine if it has a cycle in it. Follow up: Can you solve it without using extra space?
Analysis: typical Runner Technique. 一次过
1 /** 2 * Definition for singly-linked list. 3 * class ListNode { 4 * int val; 5 * ListNode next; 6 * ListNode(int x) { 7 * val = x; 8 * next = null; 9 * } 10 * } 11 */ 12 public class Solution { 13 public boolean hasCycle(ListNode head) { 14 if (head == null) return false; 15 ListNode current = head; 16 ListNode runner = head; 17 while (runner != null && runner.next != null) { 18 runner = runner.next.next; 19 current = current.next; 20 if (runner == current) return true; 21 } 22 return false; 23 } 24 }