Read N Characters Given Read4 II - Call multiple times

Design, Array, String

Hard

Given a file and assume that you can only read the file using a given method read4, implement a methodreadto read_n_characters.Your method readmay 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 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 for read.

2. The read function 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 buf is called by read.

Analysis

是同名问题的follow-up。感觉上不难，但是写起来让人抓狂。感觉画图来模拟这个过程会更好。

最终代码不长，难点在于理解这个题目究竟要做什么，并且用比较好的方法来应对多次call时，前次call用了超出需要的长度，但是由于file reader只能向前，因此需要在内部buffer保存前次read4()的结果，并且处理好在内部buffer的部分未被读取、使用的部分，用于下一次call。由于这些数字的相对大小关系不确定，如何在各种情况下操作是本题的难点。

有几个变量：

ibuf - 内部buffer，长度为read4()默认读取的长度，即4

bufCount - 代表内部临时buffer中未被复制读取过的长度，如果为0，则需要读新数据read4()，需重置bufCount = read4()，

bufEnd - 代表上次读取read4()之后，数据的长度即ibuf[bufEnd - 1]是上次读取后ibuf[]中最右末尾位置的数据

bufPtr - 记录上次拷贝ibuf[]到结果buf[]之后，ibuf[]中未读（未复制或者未使用）数据的最左端位置。

curPtr - 记录当前已拷贝到buf[]中的位置，因此外层循环的条件（之一）就是curPtr < n

Solution

根据思路实现即可。

其中

// EOF
if (bufCount == 0) {
    break;
}

是为了判断read4()已经读到文件尾。

实现：

/**
 * The read4 API is defined in the parent class Reader4.
 *     int read4(char[] buf); 
 */
public class Solution extends Reader4 {
    private int SIZE = 4;
    private int bufCount = 0;
    private int bufEnd = 0;
    private int bufPtr = 0;
    private char[] ibuf = new char[SIZE];
    /**
     * @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) {
        int curPtr = 0;
        while (curPtr < n) {

            if (bufCount == 0) {
                bufCount = read4(ibuf);
                bufEnd = bufCount;
                bufPtr = 0;
            }
            // EOF
            if (bufCount == 0) {
                break;
            }

            while (bufPtr < bufEnd && curPtr < n) {
                buf[curPtr++] = ibuf[bufPtr++];
                bufCount--;
            }
        }
        return curPtr;
    }
}

相同思路，加上EOF标记更直观

/**
 * The read4 API is defined in the parent class Reader4.
 *     int read4(char[] buf); 
 */
public class Solution extends Reader4 {
    private int SIZE = 4;
    private int bufCount = 0;
    private int bufEnd = 0;
    private int bufPtr = 0;
    private char[] ibuf = new char[SIZE];
    /**
     * @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) {
        int curPtr = 0;
        boolean EOF = false;
        while (curPtr < n && !EOF) {

            if (bufCount == 0) {
                bufCount = read4(ibuf);
                bufEnd = bufCount;
                bufPtr = 0;
                if (bufCount < 4) {
                    EOF = true;
                }
            }


            while (bufPtr < bufEnd && curPtr < n) {
                buf[curPtr++] = ibuf[bufPtr++];
                bufCount--;
            }
        }
        return curPtr;
    }
}

更简洁（短）的写法：

public class Solution extends Reader4 {
    /**
     * @param buf Destination buffer
     * @param n   Number of characters to read
     * @return    The number of actual 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);
            }
            if (buffCnt == 0) break;
            while (ptr < n && buffPtr < buffCnt) {
                buf[ptr++] = buff[buffPtr++];
            }
            if (buffPtr >= buffCnt) buffPtr = 0;
        }
        return ptr;
    }
}

Reference

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

http://www.cnblogs.com/grandyang/p/5181672.html?spm=a2c4e.11153940.blogcont346669.8.32204591TxnAxM