原题链接在这里：https://leetcode.com/problems/read-n-characters-given-read4/
题目：
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.
题解：
Given API read4, 一次最多可以read 4个char, 并把这些char保留在temp Buff中.
Ask to design another API, 能一次最多读n个char. 每次call read4, 若是返回小于4, 说明已经到了end of file, 下一次跳出while loop.
readCount计数当前共读了多少char, n-readCount就是还需要读多少个char.
Time Complexity: O(n). Space: O(1).
AC Java:
/* 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 int read(char[] buf, int n) {
boolean eof = false;
char [] temp = new char[4];
int readCount = 0;
while(!eof && readCount<n){
int count = read4(temp); //read4 API读的byte都存在temp Buff中
eof = count < 4; count = Math.min(count, n-readCount); //如果n减掉已经读的char的数目比count小，就只读n-readCount个char
for(int i = 0; i<count; i++){
buf[readCount++] = temp[i];
}
}
return readCount;
}
}