LeetCode 158: Read N Characters Given Read4 II - Call multiple times

https://leetcode.com/problems/read-n-characters-given-read4-ii-call-multiple-times/

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.

Notes: 

This question asks to use the read function multiple times. Thus, we need to record the position of previous calls. The solution is to use build-in variables: readPos is the current reading position; writePos is the position from the read4 API; and a buff[4] array to record the chars read previously.

The key logic are listed below,

1): when readPos == writePos, it is time to operate another call of read4 API;

2): if not, meaning that there are still some char left in previous call. Then we should start with the buff[4] array;

3): when writePos == 0, it means no more char left, so we should stop and return the actual number of chars read so far;

read4(buff) means store the chars read into buff. buff here serves as a cache from previous read4. Every time for a new call of read(), we need to check if there are chars remaining in the buff.

See the code below,

class Solution {
public:
    int read(char *buf, int n) {
        for (int i = 0; i < n; ++i) {
            if (readPos == writePos) {
                writePos = read4(buff);
                readPos = 0;
                if (writePos == 0) return i;
            }
            buf[i] = buff[readPos++];
        }
        return n;
    }
private:
    int readPos = 0, writePos = 0;
    char buff[4];
};