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# 0158. Read N Characters Given Read4 II - Call multiple times

<https://leetcode.com/problems/read-n-characters-given-read4-ii-call-multiple-times>

## Description

Given a `file` and assume that you can only read the file using a given method `read4`, implement a method `read` to read `n` characters. Your method `read` may be **called multiple times**.

**Method read4:**

The API `read4` reads **four consecutive characters** from `file`, then writes those characters into the buffer array `buf4`.

The return value is the number of actual characters read.

Note that `read4()` has its own file pointer, much like `FILE *fp` in C.

**Definition of read4:**

```
    Parameter:  char[] buf4
    Returns:    int
buf4[] is a destination, not a source. The results from read4 will be copied to buf4[].
```

Below is a high-level example of how `read4` works:

![](https://assets.leetcode.com/uploads/2020/07/01/157_example.png)

```
File file("abcde"); // File is "abcde", initially file pointer (fp) points to 'a'
char[] buf4 = new char[4]; // Create buffer with enough space to store characters
read4(buf4); // read4 returns 4. Now buf4 = "abcd", fp points to 'e'
read4(buf4); // read4 returns 1. Now buf4 = "e", fp points to end of file
read4(buf4); // read4 returns 0. Now buf4 = "", fp points to end of file
```

**Method read:**

By using the `read4` method, implement the method read that reads `n` characters from `file` and store it in the buffer array `buf`. Consider that you cannot manipulate `file` directly.

The return value is the number of actual characters read.

**Definition of read:**

```
    Parameters:	char[] buf, int n
    Returns:	int
buf[] is a destination, not a source. You will need to write the results to buf[].
```

**Note:**

* Consider that you cannot manipulate the file directly. The file is only accessible for `read4` but not for `read`.
* The read function may be **called multiple times**.
* Please remember to **RESET** your class variables declared in Solution, as static/class variables are persisted across multiple test cases. Please see [here](https://leetcode.com/faq/) for more details.
* You may assume the destination buffer array, `buf`, is guaranteed to have enough space for storing `n` characters.
* It is guaranteed that in a given test case the same buffer `buf` is called by `read`.

**Example 1:**

```
**Input:** file = "abc", queries = [1,2,1]
**Output:** [1,2,0]
**Explanation:** The test case represents the following scenario:
File file("abc");
Solution sol;
sol.read(buf, 1); // After calling your read method, buf should contain "a". We read a total of 1 character from the file, so return 1.
sol.read(buf, 2); // Now buf should contain "bc". We read a total of 2 characters from the file, so return 2.
sol.read(buf, 1); // We have reached the end of file, no more characters can be read. So return 0.
Assume buf is allocated and guaranteed to have enough space for storing all characters from the file.
```

**Example 2:**

```
**Input:** file = "abc", queries = [4,1]
**Output:** [3,0]
**Explanation:** The test case represents the following scenario:
File file("abc");
Solution sol;
sol.read(buf, 4); // After calling your read method, buf should contain "abc". We read a total of 3 characters from the file, so return 3.
sol.read(buf, 1); // We have reached the end of file, no more characters can be read. So return 0.
```

**Constraints:**

* `1 <= file.length <= 500`
* `file` consist of English letters and digits.
* `1 <= queries.length <= 10`
* `1 <= queries[i] <= 500`

## ac: Using Queue

```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 Queue<Character> q = new LinkedList<>();

    public int read(char[] buf, int n) {
        char[] tmp = new char[4];
        int i = 0;

        while (i < n && !q.isEmpty()) {
            buf[i++] = q.poll();
        }

        while (i < n) {
            int cnt = read4(tmp);
            if (cnt == 0) break;
            for (int j = 0;j < cnt; j++) {
                if (i < n) {
                    buf[i++] = tmp[j]; 
                } else {
                    q.offer(tmp[j]);  // store remaing in a queue
                }
            }
        }

        return i;
    }
}
```