Read N Characters Given Read4

Given a file and assume that you can only read the file using a given method read4, implement a methodreadto read_n_characters.Your methodreadmay be called multiple times.

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 youcannotmanipulate 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

Example 1:

File file("abc");
Solution sol;
// Assume buf is allocated and guaranteed to have enough space for storing all characters from the file.
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.

Example 2:

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.

Note:

  1. Consider that you cannot manipulate the file directly, the file is only accesible for read4 but not forread.

  2. Thereadfunction may be called multiple times.

  3. Please remember to RESET your class variables declared in Solution, as static/class variables are persisted across multiple test cases. Please see here for more details.

  4. You may assume the destination buffer array, buf, is guaranteed to have enough space for storing n characters.

  5. It is guaranteed that in a given test case the same buffer bufis called byread.

Note

与之前不同,这次需要把变量全局化,buffer是read4读的结果

head和tail分别对应read4的结果的头尾指针,会保留上次call的结果

空:head == tail -> 重置,走到头了;tail是0那么说明读完了直接退出

RUN: "abc", [1,2,1]

  • 第一次:tail = 3,n = 1, head = 1

  • 第二次:不enqueue,head = 1 n = 2 tail = 3

  • 第三次:head = tail = 3 then head= 0 but tail = 0 break and finish

Code

/**
 * 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   Number of characters to read
     * @return    The number of actual characters read
     */
    char[] buffer = new char[4];
    int head = 0; int tail = 0;

    public int read(char[] buf, int n) {
        int i = 0;
        while (i < n) {
            if (head == tail) { // queue is empty
                head = 0;
                tail = read4(buffer); // enqueue
                if (tail == 0) {
                    break;
                }
            }

            while (i < n && head < tail) {
                buf[i++] = buffer[head++]; // dequeue
            }
        }

        return i;
    }
}

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