Linked List Cycle - LeetCode
Given a linked list, determine if it has a cycle in it.
Follow up:
Can you solve it without using extra space?
思路:维护两个指针,一快一慢,判断两个指针能否相遇。
class Solution {
public:
bool hasCycle(ListNode *head) {
if (head == NULL) return false;
ListNode *slow = head;
if (head->next == NULL) return false;
ListNode *fast = head->next;
while (slow != fast)
{
if (slow != NULL)
slow = slow->next;
if (fast != NULL)
fast = fast->next;
if (fast != NULL)
fast = fast->next;
}
return slow != NULL;
}
};
Linked List Cycle - LeetCode的更多相关文章
- 141. Linked List Cycle - LeetCode
Question 141. Linked List Cycle Solution 题目大意:给一个链表,判断是否存在循环,最好不要使用额外空间 思路:定义一个假节点fakeNext,遍历这个链表,判断 ...
- Linked List Cycle——LeetCode
Given a linked list, determine if it has a cycle in it. Follow up:Can you solve it without using ext ...
- Linked List Cycle leetcode II java (寻找链表环的入口)
题目: Given a linked list, return the node where the cycle begins. If there is no cycle, return null. ...
- Linked List Cycle leetcode java (链表检测环)
题目: Given a linked list, determine if it has a cycle in it. Follow up: Can you solve it without usin ...
- [LeetCode] Linked List Cycle II 单链表中的环之二
Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Foll ...
- [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 ...
- 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 ...
- LeetCode解题报告:Linked List Cycle && Linked List Cycle II
LeetCode解题报告:Linked List Cycle && Linked List Cycle II 1题目 Linked List Cycle Given a linked ...
- [LeetCode] 141&142 Linked List Cycle I & II
Problem: Given a linked list, determine if it has a cycle in it. Follow up:Can you solve it without ...
随机推荐
- linux学习(二) -- ubuntu下lnmp环境的配置
亲测的教程,,希望能对大家提供些许帮助,转载请注明出处 ubuntu+nginx+mysql+php7 一.安装Nginx 1.首先添加nginx_signing.key(必须,否则出错) $ wge ...
- Careercup - Microsoft面试题 - 4639756264669184
2014-05-12 06:42 题目链接 原题: Write your own regular expression parser for following condition: az*b can ...
- IOS开发学习笔记018- 一般控件的使用
1.移动 2.动画 3.缩放 3.旋转 4.简化代码 5.总结 UIButton 的两种状态 normal highlighted 1.移动 OC语法规定:不允许直接修改某个对象中结构体属性的成员. ...
- MOCTF-火眼金睛
MOCTF-火眼金睛 http://119.23.73.3:5001/web10/ 把这个题目当作python爬虫来练习. 首先要获取到文本框里面的全部信息, import requests impo ...
- PostgreSQL drop database 显示会话没有关闭 [已解决]
错误重现 有时候需要删除某个数据库时,会报如下错误,显示有一个连接正在使用数据库,无法删除 ERROR: database "pilot" is being accessed by ...
- mybitis中对象字段与表中字段名称不匹配(复制)
开发中,实体类中的属性名和对应的表中的字段名不一定都是完全相同的,这样可能会导致用实体类接收返回的结果时导致查询到的结果无法映射到实体类的属性中,那么该如何解决这种字段名和实体类属性名不相同的冲突呢? ...
- Unity 脚本<1>
RaycastHit2D hit = Physics2D.Linecast(targetPosition, targetPosition + new Vector2(x, y)); 猜测是lineca ...
- Python机器学习数据挖掘工具sklearn安装和使用
python借助pip安装第三方库,所以首先确保电脑上已成功安装了pip. 安装sklearn前需要先安装numpy.scipy和pandas等库.安装的方式有两种: 一.前往python的组件库页( ...
- tcp协议 tcpip协议 http协议,IP,DNS,端口号
每当看到HTTP协议,tcp/ip协议,TCP协议总是蒙圈:在这里先记录一下,方面自己查看 TCP协议:TCP(Transmission Control Protocol 传输控制协议)是一种面向连接 ...
- HDU 5322 Hope ——NTT 分治 递推
发现可以推出递推式.(并不会) 然后化简一下,稍有常识的人都能看出这是一个NTT+分治的情况. 然而还有更巧妙的方法,直接化简一下递推就可以了. 太过巧妙,此处不表,建议大家找到那篇博客. 自行抄写 ...