157. Read N Characters Given Read4

Easy
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downvote有300多条，我觉得应该是题意太confusing了。我一开始以为buf是要读的文件，后来想想不太对。读了一下@param, 原来buf是Destination buffer，学过IO应该才会懂，这个是让我们把file读到buffer里面去。所以我们读的时候要写到buf里，同时要返回读过的character数目。我刚开始还是很confused, 心想既然都告诉了我们读n个，那直接返回n不就好了。然而文件总char数可能是小于n的，这种情况就不能直接返回n.
几个要点：

• 这题没有给input file, 要抽象读取这个过程，把重点放到计算读过的字符数和写入buf这里。
• 每次读4个，并且保存到一个临时的char array内，然后再复制到buf里。注意这里复制的长度不能直接写4. 有几种情况需要考虑：

1. 不需要读完这4个字符串就已经达到总数n了，那么我们只需要复制n - i个就行，所以我们这里是复制Math.min(len, n - i)个；
2. len < 4, 说明这时候其实已经读到文件尾部了，这时候copy len个就好，而且这种情况我们其实已经可以直接返回了，读的字符数为i + len个。这个时候还得考虑是不是已经读了n个，如果已经到了n个，就直接返回n,所以我们返回的是Math.min(n, i + len).

/* 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) {
        //possible cases:
        //there's more than n characters in the file, return n
        //there's less than n characters in the file, use API to determine how many we read
        for (int i = 0; i < n; i += 4){
            char[] temp = new char[4];
            int len = read4(temp);
            System.arraycopy(temp, 0, buf, i, Math.min(len, n - i));
            if (len < 4){
                return Math.min(i + len, n);
            }
        }
        return n;
    }
}