LeetCode 157. Read N Characters Given Read4

原题链接在这里：https://leetcode.com/problems/read-n-characters-given-read4/

题目：

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.

题解：

Given API read4, 一次最多可以read 4个char, 并把这些char保留在temp Buff中.

Ask to design another API, 能一次最多读n个char. 每次call read4, 若是返回小于4, 说明已经到了end of file, 下一次跳出while loop.

readCount计数当前共读了多少char, n-readCount就是还需要读多少个char.

Time Complexity: O(n). Space: O(1).

AC Java:

/**
 * 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
     */
    public int read(char[] buf, int n) {
        int readSum = 0;
        boolean eof = false;
        char[] temp = new char[4];

        while(!eof && readSum < n){
            int count = read4(temp);
            eof = count < 4;
            
            int rest = n - readSum;
            int i = 0;
            while(i < count && i < rest){
                buf[readSum++] = temp[i++];
            }
        }

        return readSum;
    }
}

跟上Read N Characters Given Read4 II - Call multiple times.