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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這道題跟I不一樣在於，read函數可能多次調用，比如read(buf,23)之后又read(buf, 25), 第一次調用時的buffer還沒用完，還剩一個char在buffer里，第二次拿出來接着用，這樣才能保證接着上次讀的地方繼續往下讀。

1. 所以應該設置這4個char的buffer為instance variable（實例變量），這樣每次調用read函數后buffer還剩的元素可以被保存下來，供給下一次讀取

2. 那么下一次調用read函數時候，怎么知道上一次buffer里面還剩不剩未讀元素呢？我們有oneRead(一次讀到buffer里的char數)，actRead(實際被讀取的char數)，oneRead-actRead就是還剩在buffer里的char數。通常oneRead == actRead, 只有當n-haveRead < oneRead時，才不等，這就是上一次調用read結束的時候。所以只要調用read函數發現oneRead != 0 時，就說明上一次調用read還剩了元素在buffer里，先讀完這個，再調用read4讀新的。oneRead也需要是instance varaible

3. 還需要設置一個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--->

上圖所示為一次read最后的情況，offset部分其實就是actRead的部分，oneRead = oneRead - actRead, 就剩下了右邊一部分在buffer里沒有讀，下一次read函數調用，發現oneRead>0, 說明上一次read還剩了一部分沒有讀。oneRead表示的其實就是上一次讀剩下的char數，offset表示這一次讀應該開始的位置

其實上圖的oneRead不一定會充滿整個右邊部分的，有可能上一次讀oneRead根本沒有讀滿整個buffer。 所以oneRead+offset並不一定等於整個buffer。這也就是為什么我們一定要用兩個變量oneRead\offset的原因，因為oneRead並不一定=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;
     }
}