[LeetCode157]Read N Characters Given Read4

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.

这道题读懂题目，花了好多时间：其实，题目让我们实现的功能就是通过他本来给你的这个read4(char buf) function 去实现一个 read(char* buf, int n)* function. 这个方程干什么的？ 就是说我有一个file, 和另一个空白地儿 (let’s say it char* buf). read4将一次读4个char 并且返回它读了几个，到我们的空白空间里去， 但如果最后， file不足4个char， 它就返回真实的char的个数。 所以利用这个函数， 我们要实现一个以此可以读n个char到我们需要的空白空间去的函数。这题简单（。。吗？）之处就在于我们实现的新函数只被call了一次，就完事了。所以actually, there are two special situations:

case 1: file size = 50, n = 23. That means, 我们只需要23个char，然而file有50个，所以最后一次即使read4会读四个char到它自己的buffer里，我们不能取所有的，只copy它前三个char到我们的buffer里就可以了！所以我们得数着到底读了多少还剩多少。

case2：file size = 14, n = 23. That means, 我们即使可以读23个，但actually没有那么多char 让你读， 所以当 read4 < 4时，就可以停了！

看到一个博主的code, 利用oneRead( from read4), haveRead(total size we have read) and actualRead( the actual number we read from tmpBuf) 可以让code更easyunderstand一点。。

oneRead = read4(tmpBuf);
actualRead = min(n - haveRead, oneRead);
haveRead += actualRead;

最后附上code：

int read(char *buf, int n) {
char* tmpBuf = new char[4];
int haveRead = 0, oneRead = 0, actualRead = 0;
while(haveRead < n){
oneRead = read4(tmpBuf);
if(oneRead < 4) break;
actualRead = min(n-haveRead, oneRead);
for(int i = 0; i < actualRead; ++i){
buf[haveRead + i] = tmpBuf[i];
}
haveRead += actualRead;
}
if(haveRead == n) return haveRead;
else if(oneRead < 4){
actualRead = min(n-haveRead, oneRead);
for(int i = 0; i < actualRead; ++i){
buf[haveRead + i] = tmpBuf[i];
}
haveRead += actualRead;
return haveRead;
}
}

或者用一个flag简化code：

int read(char *buf, int n) {
char* tmpBuf = new char[4];
int haveRead = 0;
bool lessThan4 = false;
while(!lessThan4 && haveRead < n){
int oneRead = read4(tmpBuf);
if(oneRead < 4) lessThan4 = true;
int actualRead = min(n-haveRead, oneRead);
for(int i = 0; i<actualRead; ++i){
buf[haveRead + i] = tmpBuf[i];
}
haveRead += actualRead;
}
return haveRead;
}