[LeetCode] 158. Read N Characters Given Read4 II

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 4 consecutive characters from the file, then writes those characters into the buffer array buf.

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

Note: buf4[] is destination not source, the results from read4 will be copied to buf4[]

Below is a high level example of how read4 works:

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

Method read:

By using the read4 method, implement the method read that reads n characters 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:

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:

• Consider that you cannot manipulate the file directly, the file is only accesible 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 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.

用 Read4 读取 N 个字符 II。

这个题跟版本一的区别是版本一每次call这个API的时候，read函数只跑一次；但是这道题里面，read函数有可能跑多次。举个例子，文件是'abcdefg'，多次read的时候，比如第一次n给1，那buf是‘a’，再read一次，n给2，那’a’已经读过了，所以现在buf是’bc’了， 如果再来个n=3的话，buf就是‘def’。总之就是一个test case 中read函数可以调用一次和调用多次的区别。

既然是有可能会跑多次，那么就需要用全局变量记录一些pointer，否则buf的写入就会有错。其他部分跟版本一很接近。

时间O(n)

空间O(1) - 只用了一个长度固定的额外数组

Java实现

/**
 * The read4 API is defined in the parent class Reader4.
 *     int read4(char[] buf4); 
 */

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 pointer = 0;
    private int len = 0;
    private char[] temp = new char[4];

    public int read(char[] buf, int n) {
        int index = 0;
        while (index < n) {
            if (pointer == 0) {
                len = read4(temp);
            }
            if (len == 0) {
                break;
            }
            while (index < n && pointer < len) {
                buf[index] = temp[pointer];
                index++;
                pointer++;
            }
            if (pointer >= len) {
                pointer = 0;
            }
        }
        return index;
    }
}

LeetCode 题目总结