public boolean hasCycle(ListNode head)
{
ListNode slow=head;
ListNode fast=head;
if(head==null)return false;
while(fast.next!=null)
{
if(fast.next.next==null)return false;
slow=slow.next;
fast=fast.next.next;
if(slow==fast)return true;
}
return false; }

Linked List Cycle (java)的更多相关文章

  1. leetcode 141. Linked List Cycle ----- java

    Given a linked list, determine if it has a cycle in it. Follow up:Can you solve it without using ext ...

  2. Java for LeetCode 142 Linked List Cycle II

    Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Foll ...

  3. [LeetCode] Linked List Cycle 单链表中的环

    Given a linked list, determine if it has a cycle in it. Follow up: Can you solve it without using ex ...

  4. [算法][LeetCode]Linked List Cycle & Linked List Cycle II——单链表中的环

    题目要求 Linked List Cycle Given a linked list, determine if it has a cycle in it. Follow up: Can you so ...

  5. 单链表反转的递归实现(Reversing a Linked List in Java, recursively)

    转自Reversing a Linked List in Java, recursively There's code in one reply that spells it out, but you ...

  6. 【LeetCode练习题】Linked List Cycle II

    Linked List Cycle Given a linked list, determine if it has a cycle in it. Follow up:Can you solve it ...

  7. [LintCode] Linked List Cycle(带环链表)

    描述 给定一个链表,判断它是否有环. 样例 给出 -21->10->4->5, tail connects to node index 1,返回 true. 这里解释下,题目的意思, ...

  8. 【LeetCode】142. Linked List Cycle II

    Difficulty:medium  More:[目录]LeetCode Java实现 Description Given a linked list, return the node where t ...

  9. LeetCode: Linked List Cycle II 解题报告

    Linked List Cycle II Given a linked list, return the node where the cycle begins. If there is no cyc ...

随机推荐

  1. oracle行号排序问题

    1.创建一个student,并且插入数据 ),age int) '); '); '); '); '); commit; 2.直接按照age进行排序显示行号: select * from(select ...

  2. 对list代理扩展功能

    package 动态代理扩展List; import java.lang.reflect.InvocationHandler; import java.lang.reflect.Method; imp ...

  3. jquery动态添加DOM节点

    1.append()方法:向每个匹配的元素内部添加元素 appendTo()方法:将所有匹配的元素追加的指定的元素中 <html> <head> <meta http-e ...

  4. 通过AJAX和PHP,提交JQuery Mobile表单

    File name: callajax.php <?php $firstName = $_POST[firstName]; $lastName = $_POST[lastName]; echo( ...

  5. 写入和读取LOB类型的对象

    ====写入数据============ create or replace procedure addWaterFallis directions clob; amount binary_integ ...

  6. 简易的WPF MVVM模式开发

    Model层 public class Song { private string _artistName; private string _songTitle; public string Song ...

  7. 使用meta来控制浏览器的渲染方式

    <meta name="renderer" content="webkit"><!--默认使用webkit内核(360急速模式)--> ...

  8. SQL Server 2005中的分区表(五):添加一个分区

    所谓天下大事,分久必合,合久必分,对于分区表而言也一样.前面我们介绍过如何删除(合并)分区表中的一个分区,下面我们介绍一下如何为分区表添加一个分区. 为分区表添加一个分区,这种情况是时常会 发生的.比 ...

  9. 几个js的linq实现

    几个js的linq实现 linqjs.codeplex.com jslinq.codeplex.com javascriptiqueryable.codeplex.com fromjs.codeple ...

  10. VC++6.0出现no compile tool is associated with the extension.解决方法

    对于刚解除VC++6.0的小白,在编译时候经常出现下图的错误提示: 解释为:不能编译此BmpRot.h文件,没有合适的编译工具可以编译此扩展名的文件. 很明显,当然只有.cpp文件才能编译. .h头文 ...