leetcode158

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.

 

 

和上一题比较像，区别是每次读的东西不是独立的，是和上一次读走多少个相关的。比如对123456789你读了三次，分别要6, 1, 4个，那你返回的应该是123456, 7, 89。

主要要解决的问题有两个：读多的怎么留给下次用；读到末尾时文件不够我需求怎么办。

前者通过FIFO队列（或模拟队列）解决。后者通过尽量压榨你再去读，但真的压榨到读出0个的时候就无情走人了来解决。

 

缓冲区逻辑：要新字符的时候，如果缓冲区还没空就直接拿走一个，如果缓冲区空了你就先populate一下缓冲区再接着拿走一个。

缓冲区可以用一个char[4]数组+head tail指针来模拟，那就不用又开一个queue又用char[] buf4去取4了。 

 

我的实现

/* 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
     */
    
    char[] buffer = new char[4];
    int head = 0, tail = 0;
    
    public int read(char[] buf, int n) {
        int sum = 0;
        while (sum < n) {
            if (head == tail) {
                head = 0;
                tail = read4(buffer);
                if (tail == 0) {
                    break;
                }
            }
            buf[sum++] = buffer[head++];
        }
        return sum;
    }
}