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.

* @param buf Destination buffer，我后来理解这个应该是个足够大的buffer，需要将读到的file放到这个buffer里面去
* @param n Maximum number of characters to read，想要读n个char，但是有可能读不到那么多，因为有可能n比文件的size大
* @return The number of characters read，实际读到的char数，小于等于n

由于不知道文件的大小，所以只能一次一次用4个char的buffer用read4去读，存在以下两种结束方式（假设文件大小为size）：

1. n >> size, 比如n = 50, size = 23, 最后一次读的大小为<4, 下一次就不读了，然后把这一次读的char数oneRead赋给目标buf

至于n = 50, size = 20, 最后一次读到0个，不会进行赋值，下一次也不读了

2. n << size，比如n = 23, size = 50, read4每次会读满4个，但是我们只取3个，取零头的方法是int actRead = Math.min(n-haveRead, oneRead); 下一次OneRead+haveRead>n, 也就停止了

想清楚了这些，就可以写了

 1 /* The read4 API is defined in the parent class Reader4. 2       int read4(char[] buf); */ 3  4 public class Solution extends Reader4 { 5     /** 6      * @param buf Destination buffer 7      * @param n   Maximum number of characters to read 8      * @return    The number of characters read 9      */10     public int read(char[] buf, int n) {11        char[] buffer = new char[4];12        int haveRead = 0;13        boolean lessthan4 = false;14        15        while (!lessthan4 && haveRead < n) {16            int oneRead = read4(buffer);17            if (oneRead < 4) lessthan4 = true;18            int actRead = Math.min(n-haveRead, oneRead);19            while (int i=1; i<=actRead; i++) {20                buf[haveRead+i-1] = buffer[i-1];21            }22            haveRead += actRead;23        }24        return haveRead;25     }26 }