[LeetCode158]Read N Characters Given Read4 II - Call multiple times

ac59

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.

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Hide Similar Problems (E) Read N Characters Given Read4

这道题跟之前那道几乎是一样的，但现在问题是：readN函数会被call很多次，那就意味着，如果buffer里存的char上次并没有用完，下次call的时候必须先把buffer里的读完才可以开始读新的char。

比如我们的file是字母表，第一次我们call了read(buf, 10) 我们读了 字母a – g到我们的buf里，但其实最后一次对于read4来说，它的tmpBuf里读了 i g k l, 所以如果下次我们call 了read(buf, 16): 读a到p。我们其实是在上一次的基础上再读入tmpBuf剩下的char然后再利用read4开始读新的。

所以怎么实现呢？

把oneRead 和 tmpBuffer 当作 class variable就可以了。 同时我们需要一个offset variable 来记录上次到底读到哪个char了。

note: oneRead != 0 indicates whether there was some char unread by last call readN.

int read4(char *buf);

class Solution {
public:
/**
* @param buf Destination buffer
* @param n   Maximum number of characters to read
* @return    The number of characters read
*/
char buffer[4];
int oneRead = 0;
int offset = 0;
int read(char *buf, int n) {
bool lessThan4 = false;
int haveRead = 0;
while (!lessThan4 && haveRead < n ) {
if(!oneRead){// no leftover.
oneRead = read4(buffer);
lessThan4 = (oneRead < 4);
}
int actualRead = min(n-haveRead , oneRead);
for(int i = 0; i<actualRead; ++i){//copy char from buffer to target buf!
buf[haveRead+i] = buffer[offset+i];
}
oneRead -= actualRead; // if there are some leftover, then oneread should be large than 0;
offset = (offset + actualRead) %4;
haveRead += actualRead;
}
return haveRead;
}
};