158. Read N Characters Given Read4 II - Call multiple times.java

/* The read4 API is defined in the parent class Reader4.
      int read4(char[] buf); */
## Idea 
* Based on 159
* Because of call multiple times, there are more corner cases
* First time we call, we need to save extra characters for next call
*  In the while loop, if buffPointer reaches current buffCount, it will be set as zero to be ready to read new data.
public class Solution extends Reader4 {
    /**
     * @param buf Destination buffer
     * @param n   Maximum number of characters to read
     * @return    The number of characters read
     */
    private int pointer = 0;  //  temp数组的下标
    private int counter = 0;  // temp中的元素个数
    private char[] temp = new char[4]; //can be 4096 here in interview
  
    //重点理解：temp数组：pointer的上界是counter； buf数组：index的上界是n
    public int read(char[] buf, int n) {
        int index = 0;// buf的下标，也就是我们所有已经读完的个数。index of buf, also means the total chars we have read
        while (index < n) {//注意整个循环中index是持续增长的，没有清零过程，意味着新读入的字符一直往后加
            if (pointer == 0) {//this means we should read new chars  该读新的了
                counter = read4(temp);
            }
            if (counter == 0) {//if no more chars can be read (n > total chars in file) 没有可以读的了，break
                break;
            }
            //buf和temp两数组中，指针都没到最后时，赋值。执行次数看这两边哪个距离小执行几次
            while (index < n && pointer < counter) {//pointer < counter, not <= cuz counter is num of chars, not index
                buf[index++] = temp[pointer++];//both pointers should be increased
            }
            //temp读完到头了，pointer清零
            if (pointer == counter) {//if all chars in temp have been read, update the pointer to 0 (need read new chars)
                pointer = 0;
            }
        }
        return index;
    }
}