Leetcode: Read N Characters Given Read4 II - Call multiple times

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.

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这道题跟I不一样在于，read函数可能多次调用，比如read(buf,23)之后又read(buf, 25), 第一次调用时的buffer还没用完，还剩一个char在buffer里，第二次拿出来接着用，这样才能保证接着上次读的地方继续往下读。

1. 所以应该设置这4个char的buffer为instance variable（实例变量），这样每次调用read函数后buffer还剩的元素可以被保存下来，供给下一次读取

2. 那么下一次调用read函数时候，怎么知道上一次buffer里面还剩不剩未读元素呢？我们有oneRead(一次读到buffer里的char数)，actRead(实际被读取的char数)，oneRead-actRead就是还剩在buffer里的char数。通常oneRead == actRead, 只有当n-haveRead < oneRead时，才不等，这就是上一次调用read结束的时候。所以只要调用read函数发现oneRead != 0 时，就说明上一次调用read还剩了元素在buffer里，先读完这个，再调用read4读新的。oneRead也需要是instance varaible

3. 还需要设置一个offset: Use to keep track of the offset index where the data begins in the nextread call. The buffer could be read partially (due to constraints of reading upto n bytes) and therefore leaving some data behind.

                      |<--buffer-->|   
    // |_________________________|
    // |<---offset---> |<----oneRead--->

上图所示为一次read最后的情况，offset部分其实就是actRead的部分，oneRead = oneRead - actRead, 就剩下了右边一部分在buffer里没有读，下一次read函数调用，发现oneRead>0, 说明上一次read还剩了一部分没有读。oneRead表示的其实就是上一次读剩下的char数，offset表示这一次读应该开始的位置

其实上图的oneRead不一定会充满整个右边部分的，有可能上一次读oneRead根本没有读满整个buffer。 所以oneRead+offset并不一定等于整个buffer。这也就是为什么我们一定要用两个变量oneRead\offset的原因，因为oneRead并不一定=4-offset

/* 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
     */
     private char buffer = new char[4];
     private int oneRead = 0;
     private int offset = 0;
     
     public int read(char[] buf, int n) {
         boolean lessthan4 = false;
         int haveRead = 0;
         while (!lessthan4 && haveRead < n) {
             if (oneRead == 0) {
                 oneRead = read4(buffer);
                 lessthan4 = oneRead < 4;
             }
             int actRead = Math.min(n-haveRead, oneRead);
             for (int i=0; i<actRead; i++) {
                 buf[haveRead+i] = buffer[offset+i];
             }
             oneRead -= actRead;
             offset = (offset + actRead) % 4;
             haveRead += actRead;
         }
         return haveRead;
     }
}