# Problem 158: Read N Characters Given Read4 II - Call multiple times
>
## 思路
* 这道题比上一道题难度增大，上一题说 read 函数只能调用一次，而这道题说read函数可以调用多次，为了更简单直观的说明问题，我们举个简单的例子吧。
比如：buf = "ab", \[read(1),read(2)]，返回 \["a","b"]，那么第一次调用read(1) 后，从 buf 中读出一个字符，那么就是第一个字符 a，然后又调用了一个read(2)，想取出两个字符，但是 buf 中只剩一个b了，所以就把取出的结果就是 b。
再来看一个例子：buf = "a", \[read(0),read(1),read(2)]，返回 \["","a",""]，第一次调用 read(0)，不取任何字符，返回空，第二次调用 read(1)，取一个字符，buf 中只有一个字符，取出为a，然后再调用 read(2)，想取出两个字符，但是 buf 中没有字符了，所以取出为空。
* 这里我们需要记录上一次 buff 的使用情况： buffPtr 代表 buff 的 pointer，看他指到了哪儿； buffCnt 代表的是 buff 的index，看读完了没有。
```java
/* 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 int buffPtr = 0;
private int buffCnt = 0;
private char[] buff = new char[4];
public int read(char[] buf, int n) {
int ptr = 0;
while (ptr < n) {
if (buffPtr == 0) {
buffCnt = read4(buff);
}
while (ptr < n && buffPtr < buffCnt) {
buf[ptr++] = buff[buffPtr++];
}
// all chars in buff used up, set pointer to 0
if (buffPtr == buffCnt) {
buffPtr = 0;
}
// read4 returns less than 4, end of file
if (buffCnt < 4) {
break;
}
}
return ptr;
}
}
```
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