Leetcode - Read N Characters Giv

My code:

/* The read4 API is defined in the parent class Reader4.
      int read4(char[] buf); */

public class Solution extends Reader4 {
    Queue<Character> q = new LinkedList<Character>();
    /**
     * @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) {
        char[] arr = new char[4];
        int counter = 0;
        int num = 4;
       
        while (num == 4 && counter < n) {
            while (!q.isEmpty() && counter < n) {
                buf[counter] = q.poll();
                counter++;
            }
            if (counter >= n) {
                return counter;
            }
            
            num = read4(arr);
            int rightMost = Math.min(counter + num, n);
            for (int i = counter; i < rightMost; i++) {
                buf[i] = arr[i - counter];
            }
            if (counter + num > n) {
                for (int i = n; i < counter + num; i++) {
                    q.offer(arr[i - counter]);
                }
            }
            counter = Math.min(counter + num, n);
        }
       
        return counter;
    }
}

自己做了出来。其实还好。看懂题意就简单多了。
用 Queue 来存储读多了的字符，下次用。

Anyway, Good luck, Richardo! -- 09/20/2016