Read N Characters Given Read4

https://leetcode.com/problems/read-n-characters-given-read4/description/

The API: int read4(char *buf) reads 4 characters at a time from a file.

The return value is the actual number of characters read. For example, it returns 3 if there is only 3 characters left in the file.

By using the read4 API, implement the function int read(char *buf, int n) that reads n characters from the file.

Note:

The read function will only be called once for each test case.

Thoughts

题目意思是给一个read4(char[] inBuffer)函数, 它能从某处读入4个字符写入到inBuffer里, 现在让你调用read4, 实现一个general的read(int[] buf, n)的函数, 即读入n个字符进buf里. 那我们首先需要一个internal buffer来接read4的写入, 然后设当前遍历到的位置为index, 调用read4()每次读入4个进inBuffer中, 再把buf[index]不断写入inBuffer中还没写入到buf中的. 直到read4返回0即没有可读或index == n.

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   Maximum number of characters to read
     * @return    The number of characters read
     */
    public int read(char[] buf, int n) {
        char[] inBuf = new char[4];
        int index = 0;
        while (index < n) {
            int count = read4(inBuf);
            if (count == 0) {
                break;
            }
            for (int i = 0; i < count && index < n; i++) {
                buf[index++] = inBuf[i];
            }
        }
        return index;
    }
}

Analysis

时间复杂度O(N).