leetCode题解 Student Attendance Record I
1、题目描述
You are given a string representing an attendance record for a student. The record only contains the following three characters:
- 'A' : Absent.
- 'L' : Late.
- 'P' : Present.
A student could be rewarded if his attendance record doesn't contain more than one 'A' (absent) or more than two continuous 'L' (late).
You need to return whether the student could be rewarded according to his attendance record.
Example 1:
Input: "PPALLP"
Output: True Example 2:
Input: "PPALLL"
Output: False 输入一个string ,如果其中连续出现出现两次 ‘A’,或者连续出现三次 ‘L’返回false。 2、代码
bool checkRecord(string s) {
int numA = ;
int numL = ;
for(int i = ; i < s.size(); i++)
{
if(s[i] == 'A' && ++numA > )
return false;
if(s[i] == 'L')
{
numL++;
if(numL > )
return false;
}
else
numL = ;
}
return true;
}
leetCode题解 Student Attendance Record I的更多相关文章
- [LeetCode] 552. Student Attendance Record II 学生出勤记录之二
Given a positive integer n, return the number of all possible attendance records with length n, whic ...
- LeetCode 551. Student Attendance Record I (学生出勤纪录 I)
You are given a string representing an attendance record for a student. The record only contains the ...
- LeetCode 551. Student Attendance Record I (C++)
题目: You are given a string representing an attendance record for a student. The record only contains ...
- [LeetCode] Student Attendance Record II 学生出勤记录之二
Given a positive integer n, return the number of all possible attendance records with length n, whic ...
- 【leetcode】552. Student Attendance Record II
题目如下: Given a positive integer n, return the number of all possible attendance records with length n ...
- [Swift]LeetCode552. 学生出勤记录 II | Student Attendance Record II
Given a positive integer n, return the number of all possible attendance records with length n, whic ...
- 552. Student Attendance Record II
Given a positive integer n, return the number of all possible attendance records with length n, whic ...
- 551. Student Attendance Record I 从字符串判断学生考勤
[抄题]: You are given a string representing an attendance record for a student. The record only contai ...
- 551. Student Attendance Record I【easy】
551. Student Attendance Record I[easy] You are given a string representing an attendance record for ...
随机推荐
- Aop学习笔记系列一
一.Aop解决了什么问题? 1.在说解决了什么问题之前,先介绍一些关键的知识点 a.功能需求:功能需求指项目中的增值需求,比如业务逻辑,UI,持久化(数据库). b.非功能需求:项目中次要的,但却不可 ...
- apache URL重写 标志表 以及 错误解决方法
Apache mod_rewrite规则重写的标志一览 1) R[=code](force redirect) 强制外部重定向 强制在替代字符串加上http://thishost[:thisport] ...
- Android 开发工具类 21_SAXForHandler
解析 XML 有两种形式: 1.XMLReader XMLReaser xmlReader = saxParser.getXMLReader(); xmlReadeer.setContentHandl ...
- Android_读取元素的数据
在AndroidManifest.xml中,<meta-data>元素可以作为子元素,被包含在<activity>.<application> .<servi ...
- Disconf 学习系列之全网最详细的最新稳定Disconf 搭建部署(基于Ubuntu14.04 / 16.04)(图文详解)
不多说直接上干货! https://www.cnblogs.com/wuxiaofeng/p/6882596.html (ubuntu16.04) https://www.cnblogs.com/he ...
- docker容器网络通信原理分析(转)
概述 自从docker容器出现以来,容器的网络通信就一直是大家关注的焦点,也是生产环境的迫切需求.而容器的网络通信又可以分为两大方面:单主机容器上的相互通信和跨主机的容器相互通信.而本文将分别针对这两 ...
- 不开vip会员照样看vip电影(亲测有效)
此为临时链接,仅用于文章预览,将在短期内失效关闭 不开vip会员照样看vip电影(亲测有效) 2018-03-08 mr_lee Python达人课堂 刚刚测试,真实有效,颇不接待要分享了... 土豪 ...
- PHP面向对象常见符号总结($this-> 、self ::)
转载:http://wyllife.blog.163.com/blog/static/4116390120116223528180/ 在php中常见的对象符号 1.$this this是指向当前对象的 ...
- cgroups简单使用
Cgroups控制系统资源的分配(cpu.mem.io) 1.cgroups概述 CGroup是Linux内核提供的可以限制.隔离进程组 (process groups) 所使用的物理资源 (如 cp ...
- Beta阶段——Scrum 冲刺博客第一天
一.当天站立式会议照片一张 二.每个人的工作 (有work item 的ID),并将其记录在码云项目管理中 昨天已完成的工作 今日是Beta冲刺第一天,昨日没有完成的工作 今天计划完成的工作 实现对i ...