Read N Characters Given Read4 II - Call multiple times
The API: int read4(char *buf) reads 4 characters at a time from a file.
The return value is the actual number of characters read. For example, it returns 3 if there is only 3 characters left in the file.
By using the read4 API, implement the function int read(char *buf, int n) that reads n characters from the file.
Note:
The read function may be called multiple times.
/* The read4 API is defined in the parent class Reader4.
int read4(char[] buf); */ public class Solution extends Reader4 {
/**
* @param buf Destination buffer
* @param n Maximum number of characters to read
* @return The number of characters read
*/
private char[] tmp = new char[4];
private int tmpPoint; //indicate which position we should start reading
private int tmpContain;//indicate how many char in the tmp buff
public int read(char[] buf, int n) {
int cur = 0;
while(cur < n){
if(tmpPoint == 0){
tmpContain = read4(tmp);
}
if(tmpContain == 0){
return cur;
}
while(cur < n && tmpPoint < tmpContain){
buf[cur ++] = tmp[tmpPoint ++];
}
tmpPoint = tmpPoint % tmpContain;
}
return cur;
}
}
Read N Characters Given Read4 II - Call multiple times的更多相关文章
- [Locked] Read N Characters Given Read4 & Read N Characters Given Read4 II - Call multiple times
Read N Characters Given Read4 The API: int read4(char *buf) reads 4 characters at a time from a file ...
- [LeetCode] Read N Characters Given Read4 II - Call multiple times 用Read4来读取N个字符之二 - 多次调用
The API: int read4(char *buf) reads 4 characters at a time from a file. The return value is the actu ...
- LeetCode Read N Characters Given Read4 II - Call multiple times
原题链接在这里:https://leetcode.com/problems/read-n-characters-given-read4-ii-call-multiple-times/ 题目: The ...
- 【LeetCode】158. Read N Characters Given Read4 II - Call multiple times
Difficulty: Hard More:[目录]LeetCode Java实现 Description Similar to Question [Read N Characters Given ...
- ✡ leetcode 158. Read N Characters Given Read4 II - Call multiple times 对一个文件多次调用read(157题的延伸题) --------- java
The API: int read4(char *buf) reads 4 characters at a time from a file. The return value is the actu ...
- 158. Read N Characters Given Read4 II - Call multiple times
题目: The API: int read4(char *buf) reads 4 characters at a time from a file. The return value is the ...
- [leetcode]158. Read N Characters Given Read4 II - Call multiple times 用Read4读取N个字符2 - 调用多次
The API: int read4(char *buf) reads 4 characters at a time from a file. The return value is the actu ...
- leetcode[158] Read N Characters Given Read4 II - Call multiple times
想了好一会才看懂题目意思,应该是: 这里指的可以调用更多次,是指对一个文件多次操作,也就是对于一个case进行多次的readn操作.上一题是只进行一次reandn,所以每次返回的是文件的长度或者是n, ...
- Leetcode-Read N Characters Given Read4 II
The API: int read4(char *buf) reads 4 characters at a time from a file. The return value is the actu ...
随机推荐
- 【HNOI2015】开店
题面 题解 树链剖分 + 主席树 先考虑一个简单一点的问题: [LNOI2014]LCA 我们考察\(dep[\mathrm{LCA}(i, x)]\)的性质,发现它是\(i\)和\(x\)的链交的长 ...
- 【HNOI2014】江南乐
题面 题解 知识引入 - \(SG\)函数 任何一个公平组合游戏都可以通过把每个局面看成一个顶点,对每个局面和它的子局面连一条有向边来抽象成这个"有向图游戏".下面我们就在有向无环 ...
- ModelForm解密
一.复用model表和字段 models.py文件 class User(models.Model): username = models.CharField(max_length=32) emai ...
- selenium select 选择下拉框
实战百度首页设置,浏览偏好设置. 打开首页,在非登录的情况下,查看分析页面元素,我们可以看到,我们首先要点击的是设置, 接着点击,搜索设置, 然后select选择下拉框. select_by_inde ...
- SpringBoot日记——登录与拦截器篇
之前的文章我们把登录页写了出来,但是想要让登录现实他的基本功能,要如何做呢?本篇文章就来帮你实现第一步,让登录页对账号密码做校验,并且完成登录跳转. LoginController 1. 要实现登录, ...
- [PLC]ST语言三:OUT/OUT_T/OUT_C/OUT_C-C32
一:OUT/OUT_T/OUT_C/OUT_C-C32 说明:简单的顺控指令不做其他说明. 控制要求:无 编程梯形图: 结构化编程ST语言: (*OUT(EN,D);*) ...
- 树莓派3b添加python时间同步脚本
树莓派没有电池,因此断电后系统时间会停止,直到你开机后又继续计时,所以会造成系统时间和实际时间有很大的误差. 因为项目需要用到本地时间,精度要求不高不想折腾(如果需要高精度,需要安装ntp),所以考虑 ...
- 模块化开发之butterknife 在 library中使用
在Android开发中butterknife是一个很好的对资源初始化的工具,它可以使你的代码简洁通俗易懂,同时配合Android ButterKnife Zelezny插件可以让你写代码的速度提升至少 ...
- Spring中的数据库事物管理
Spring中的数据库事物管理 只要给方法加一个@Transactional注解就可以了 例如:
- 【坚持】Selenium+Python学习之从读懂代码开始 DAY6
2018/05/23 Python内置的@property装饰器 [@property](https://www.programiz.com/python-programming/property) ...