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 will only be called once for each test case.

这题用C 写应该简单很多。

 /* 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
*/
public int read(char[] buf, int n) {
int result = 0;
char[] tmp = new char[4];
while(result < n){
int size = super.read4(tmp);
for(int i = 0; i < size && result < n; i ++, result ++){
buf[result] = tmp[i];
}
if(size < 4) break;
}
return result;
}
}

Read N Characters Given Read4的更多相关文章

  1. [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 ...

  2. [LeetCode] Read N Characters Given Read4 用Read4来读取N个字符

    The API: int read4(char *buf) reads 4 characters at a time from a file.The return value is the actua ...

  3. Read N Characters Given Read4 I & II

    The API: int read4(char *buf) reads 4 characters at a time from a file. The return value is the actu ...

  4. 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 ...

  5. LeetCode Read N Characters Given Read4

    原题链接在这里:https://leetcode.com/problems/read-n-characters-given-read4/ 题目: The API: int read4(char *bu ...

  6. [LeetCode#157] Read N Characters Given Read4

    Problem: The API: int read4(char *buf) reads 4 characters at a time from a file. The return value is ...

  7. [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 ...

  8. 【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 ...

  9. [LeetCode] Read N Characters Given Read4 I & II

    Read N Characters Given Read4 The API: int read4(char *buf) reads 4 characters at a time from a file ...

  10. [LeetCode] 157. Read N Characters Given Read4 用Read4来读取N个字符

    The API: int read4(char *buf) reads 4 characters at a time from a file.The return value is the actua ...

随机推荐

  1. 5290: [Hnoi2018]道路

    5290: [Hnoi2018]道路 链接 分析: 注意题目中说每个城市翻新一条连向它的公路或者铁路,所以两种情况分别转移一下即可. 注意压一下空间,最后的叶子节点不要要访问,空间少了一半. 代码: ...

  2. CF 1114 D. Flood Fill

    D. Flood Fill 链接 题意: 一个颜色序列,每个位置有一个颜色,选择一个起始位置,每次可以改变包含这个位置的颜色段,将这个颜色段修改为任意一个颜色, 问最少操作多少次.n<=5000 ...

  3. JZOJ 10043 第k小数

    Description 有两个非负整数数列,元素个数分别为N和M.从两个数列中分别任取一个数相乘,这样一共可以得到NM个数,询问这NM个数中第K小数是多少. 时间限制为20ms . Input 输入文 ...

  4. 【Maven】在pom.xml文件中使用resources插件的小作用

    在spring boot创建web项目打包为jar包的过程中,是不会把webapp目录下的页面也打包进去的,这个时候接触到了maven的 resources插件. ================== ...

  5. AJAX其实就是一个异步网络请求

    AJAX = Asynchronous JavaScript and XML(异步的 JavaScript 和 XML).其实就是一个异步网络请求. 一.创建对象 var xmlhttp; if (w ...

  6. 【厚积薄发】Crunch压缩图片的AssetBundle打包

    这是第133篇UWA技术知识分享的推送.今天我们继续为大家精选了若干和开发.优化相关的问题,建议阅读时间10分钟,认真读完必有收获. UWA 问答社区:answer.uwa4d.com UWA QQ群 ...

  7. play-with-vim1~5

    1.移动 h,j,k,l分别对应左下上右 2.模式 vim有四种模式:普通模式,插入模式,可视模式,命令行模式 进入vim 默认为普通模式,光标为方块 输入i 进入插入模式,窗口左下角为insert ...

  8. SQLMAP学习笔记1 access注入

    SQLMAP学习笔记1  access注入 Sqlmap是开源的自动化SQL注入工具,由Python写成,具有如下特点: 完全支持MySQL.Oracle.PostgreSQL.Microsoft S ...

  9. Redux 入门教程(一):基本用法

    转自http://www.ruanyifeng.com/blog/2016/09/redux_tutorial_part_one_basic_usages.html(仅供个人学习使用) 首先明确一点, ...

  10. Django之自带认证

    自带登录实例 {% extends "layout/base.html" %} // 所有link {% block body %} <div id="contai ...