Read N Characters Given Read4

Easy

Given a file and assume that you can only read the file using a given method read4, implement a method to read_n_characters.

Method read4:

The API read4reads 4 consecutive characters from the file, then writes those characters into the buffer arraybuf.

The return value is the number of actual characters read.

Note that read4()has its own file pointer, much likeFILE *fpin C.

Definition of read4:

    Parameter:  char[] buf
    Returns:    int

Note: buf[] is destination not source, the results from read4 will be copied to buf[]

Below is a high level example of howread4works:

File file("abcdefghijk"); // File is "abcdefghijk", initially file pointer (fp) points to 'a'
char[] buf = new char[4]; // Create buffer with enough space to store characters
read4(buf); // read4 returns 4. Now buf = "abcd", fp points to 'e'
read4(buf); // read4 returns 4. Now buf = "efgh", fp points to 'i'
read4(buf); // read4 returns 3. Now buf = "ijk", fp points to end of file

Method read:

By using theread4method, implement the method readthat readsncharacters from the file and store it in the buffer array buf. Consider that you cannot manipulate the file directly.

The return value is the number of actual characters read.

Definition of read:

    Parameters:    char[] buf, int n
    Returns:    int

Note: buf[] is destination not source, you will need to write the results to buf[]

Example 1:

Input: 
file = "abc", n = 4

Output: 
3

Explanation:
 After calling your read method, buf should contain "abc". We read a total of 3 characters from the file, so return 3. Note that "abc" is the file's content, not buf. buf is the destination buffer that you will have to write the results to.

Example 2:

Input: 
file = "abcde", n = 5

Output: 
5

Explanation: 
After calling your read method, buf should contain "abcde". We read a total of 5 characters from the file, so return 5.

Example 3:

Input: 
file = "abcdABCD1234", n = 12

Output: 
12

Explanation: 
After calling your read method, buf should contain "abcdABCD1234". We read a total of 12 characters from the file, so return 12.

Example 4:

Input: 
file = "leetcode", n = 5

Output: 
5

Explanation: 
After calling your read method, buf should contain "leetc". We read a total of 5 characters from the file, so return 5.

Note:

  1. Consider that you cannot manipulate the file directly, the file is only accesible for read4 but not for read.
  2. The read function will only be called once for each test case.
  3. You may assume the destination buffer array, buf, is guaranteed to have enough space for storing n characters.

Solution & Analysis

Source: https://mnmunknown.gitbooks.io/algorithm-notes/728,_fb_tag_ti.html

重新理一下这题的 API 和题目描述:

  • 已知 int read4(char[] buf) 可用
  • 你的程序要传入一个 char[] 过去,函数写入最多 4 个 char 在上面,然后返回写入 char 的长度; 到文件末尾的时候返回 char 长度会小于 4.

  • 于是我们的目标是把最终长度为 n 的字符串写入自己函数的 read(char[] buf) 里,然后返回实际长度。

主要的注意点和难点

有的时候 read4 是短板,有的时候 readN 自己是短板(不需要4个那么多的字符),在写入的时候要注意处理下。

可以通过

// get the actual count
count = Math.min(count, n - total);

或者

while (i < count && cur < n) {
...
}

来限制。

思路:

比较巧妙的是用一个pointer cur记录已经从临时buffer tmp[]写完到buf[]的位置,外层循环的条件则是pointer cur < n

EOF的条件则是read4() 返回的count < SIZE, 也就是 count < 4。

/**
 * The read4 API is defined in the parent class Reader4.
 *     int read4(char[] buf);
 */
public class Solution extends Reader4 {
    private int SIZE = 4;
    /**
     * @param buf Destination buffer
     * @param n   Number of characters to read
     * @return    The number of actual characters read
     */
    public int read(char[] buf, int n) {
        char[] tmp = new char[SIZE];
        int cur = 0;
        int count = 0;
        boolean EOF = false;
        while (!EOF && cur < n) {
            count = read4(tmp);
            EOF = (count < SIZE);

            int i = 0;
            while (i < count && cur < n) {
                buf[cur++] = tmp[i++];
            }
        }

        return cur;
    }
}

*Using EOF, count, tmp[]

public int read(char[] buf, int n) {
  boolean eof = false;      // end of file flag
  int total = 0;            // total bytes have read
  char[] tmp = new char[4]; // temp buffer

  while (!eof && total < n) {
    int count = read4(tmp);

    // check if it's the end of the file
    eof = count < 4;

    // get the actual count
    count = Math.min(count, n - total);

    // copy from temp buffer to buf
    for (int i = 0; i < count; i++) 
      buf[total++] = tmp[i];
  }

  return total;
}

The reason we need Math.min(count, n-total) is because we only want to read N characters even if we have all 4 characters returned from Read4.

Say, N = 18 and we're implentening Read18, then Read4 will takes us to 4*4 = 16 chars. After that, we only want to read 2 more chars (even if the next Read4 returns 3 or 4 characters).

Another Implementation

public int read(char[] buf, int n) {

    char[] temp = new char[4];  //Store our read chars from Read4
    int total = 0;

    while(total < n){
        // Read and store characters in Temp. Count will store total chars read from Read4
        int count = read4(temp);

        // Even if we read 4 chars from Read4, we don't want to exceed N and only want to read chars till N.
        count = Math.min(count, n-total);

        // Transfer all the characters read from Read4 to our buffer
        for(int i = 0;  i < count; i++){
            buf[total] = temp[i];
            total++;
        }

        // EOF. Done. We can't read more characters.
        if(count < 4) break;
    }

    return total;
}

Reference

https://mnmunknown.gitbooks.io/algorithm-notes/728,_fb_tag_ti.html

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