leetcode:Linked List Cycle

题目

Link:Linked List Cycle

Given a linked list, determine if it has a cycle in it.

Follow up:

Can you solve it without using extra space?

我的解法

这道题在2015年腾讯实习生笔试中出现。

很明显的解法是，设置两个指针，一个走两步，一个走一步。如果有环，走两步的指针肯定能追上走一步的；若没有环，走两步的肯定先走到链表尾部。

use faster and lower runner solution. (2 pointers)

the faster one move 2 steps, and slower one move only one step.

if there’s a circle, the faster one will finally “catch” the slower one.

(the distance between these 2 pointers will decrease one every time.)

if there’s no circle, the faster runner will reach the end of linked list. (NULL)

java中没有指针，只有引用。而它又和C++的引用完全不是一个东西，c/c++的引用是同一块内存的不同名字。而java的引用是指向一个对象，引用本身也占用了内存。

另外，java中，链表节点通过class实现，在C/C++中通过struct实现

我的代码：

/**
* 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 (head == null || head.next == null)
return false;
ListNode slow = head;
ListNode fast = head;
while(fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next.next;
if (slow == fast)
return true;
}
return false;
}
}