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 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.
Analysis:
Since the function can be called multiple times, we need to record the left over content in the buffer of read4, and put them in some place. For the next call, we need to read content from the carry over buffer first.
Solution:
/* 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 char[] carry;
public int index; public int read(char[] buf, int n) {
int left = n; //Put carry over into buf, and destroy the carry over array!
if (carry!=null){
while (left>0 && index<carry.length){
buf[n-left]=carry[index];
left--;
index++;
}
if (index>=carry.length){
carry=null;
index=-1;
}
} char[] tempBuf = new char[4];
while (left>0){
int num = read4(tempBuf);
//if the read number is larger then what we need, then we just put the left number of chars into buf.
//And put the rest chars into carry array.
if (num>left){
carry = new char[num-left];
for (int i=left;i<num;i++)
carry[i-left] = tempBuf[i];
index = 0;
} int end = Math.min(num,left);
for (int i=0;i<end;i++){
buf[n-left] = tempBuf[i];
left--;
} //If reach EOF.
if (left>0 && num<4) break;
} return n-left; }
}
Leetcode-Read N Characters Given Read4 II的更多相关文章
- [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] 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 ...
- [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 ...
- 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 ...
- [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】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 ...
- 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]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 ...
随机推荐
- 1166 矩阵取数游戏[区间dp+高精度]
1166 矩阵取数游戏 2007年NOIP全国联赛提高组 时间限制: 1 s 空间限制: 128000 KB 题目等级 : 黄金 Gold 题解 题目描述 Description [ ...
- 深入理解ThreadLocal(转)(2015年06月11日)
注明:转自:http://my.oschina.net/clopopo/blog/149368 学习一个东西首先要知道为什么要引入它,就是我们能用它来干什么.所以我们先来看看ThreadLocal对我 ...
- 四.CSS声明
概述 一条CSS规则包括选择符和声明两个部分,声明又包括属性和值,属性指出要影响样式的那个方面,而值用于把属性设定成什么 每个HTML元素有很多属性,可以用CSS来设定,但值却只有少数的几种 CSS属 ...
- CSS之transition(动画)
Transform字面上就是变形,改变的意思.在CSS中transform主要包括以下几种:旋转rotate.扭曲skew.缩放scale和移动translate以及矩阵变形matrix. 下面我们一 ...
- .net framework缓存遍历
背景: 公司的老框架里的登录信息用的MemoryCache保存的,为了实现单用户登录(即一个账号不能同事登录),需要在登录前对已经登录的信息做遍历. 大致思路如下: 本方法可用于清除所有的缓存. 1. ...
- OC7_代理的基本概念
// // Cat.h // OC7_代理的基本概念 // // Created by zhangxueming on 15/6/24. // Copyright (c) 2015年 zhangxue ...
- Optimistic Offline Lock乐观离线锁
通过冲突检测和(发生冲突时的)事务回滚,来防止并发业务事务中的冲突. 通常一个业务事务的执行,会跨越一系列的系统事务. 一旦超出了单个系统事务的范围,就不能仅依靠DB管理程序来保证数据一致性. 乐观离 ...
- 8款超酷的HTML5 3D图片动画源码
1.HTML5移动端图片左右切换动画 今天要给大家分享一款很不错的图片左右切换焦点图动画,并且支持移动端触摸滑动.功能上,这款HTML5图片播放器支持鼠标滑动.手机端触摸滑动以及自动播放.外观上,这款 ...
- zedboard 构建嵌入式linux
本文通过五部完成zedboard的嵌入式LINUX搭建,所谓磨刀不五砍材工嘛 1:系统环境搭建 要准备好交叉编译环境 见http://blog.csdn.net/xiabodan/article/de ...
- Apache+lvs高可用+keepalive(主从+双主模型)
Apache+lvs高可用+keepalive(主从+双主模型) keepalive实验准备环境: httpd-2.2.15-39.el6.centos.x86_64 keepalived-1 ...