Given a linked list, determine if it has a cycle in it.
Follow up:
Can you solve it without using extra space?
思路:快慢指针,设置一个走两步的快指针,和一个走一步的慢指针,如果有环,则它们一定会相遇。
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) 15 return false; 16 ListNode p1 = head,p2 = head.next; 17 while (p2!=null){ 18 p2 = p2.next; 19 if (p2==null) 20 return false; 21 p2 = p2.next; 22 p1 = p1.next; 23 if (p2==p1) 24 return true; 25 } 26 return false; 27 } 28 }