用read 实现readn... 至调用一次，感觉怎么搞都可以。。。估计这个题有II就是调用多次了。。。 感觉多次勇哥buffer存下多读的那部分就好了。。。 ...
2014-11-21 23:28 4 1940 推荐指数：
Given a file and assume that you can only read the file using a given method read4, implement a method to read n characters. Method read4 ...
这道题跟I不一样在于，read函数可能多次调用，比如read(buf,23)之后又read(buf, 25), 第一次调用时的buffer还没用完，还剩一个char在buffer里，第二次拿出来接着用，这样才能保证接着上次读的地方继续往下读。 1. 所以应该设置这4个char ...
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 ...
leetcode上面写的难度是hard，其实很简单，前面也写了，加个buffer就好了。 ...
Given a file and assume that you can only read the file using a given method read4, implement a method read to read n characters. Your method read ...
We have a sorted set of digits `D`, a non-empty subset of `{'1','2','3','4','5','6','7','8','9'}`. ...
Let's define a function `countUniqueChars(s)` that returns the number of unique characters on `s`, for example if `s = "LEETCODE"` then `"L"`, `"T ...
ssize_t read(int fildes, void *buf, size_t nbyte); 返回值: > 0: 实际读到的字节数 = 0: 读完数据(读文件, 管道, socket末尾-->对端关闭, 对端未关闭会一直等待) -1: 异常: errno ...
