✡ leetcode 157. Read N Characters Given Read4 利用read4实现read --------- java

时间:2022-06-01 20:31:23

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.

read4(char[] buf)指的是读取4个数,存储在buf中,然后返回成功读取的数目,如果不够四个,那么就返回剩余数目的字符。

实现的read(char[] buf, int n)功能类似,只不过将4改为了n,需要输入而已。(这里一个例子只会读取一次read)

刚开始做的时候理解错了,以为是读取buf中的字符,导致提交失败。

方法是:先读取n/4次read4,如果期间有出现了不等于4的情况,那么返回结果。

    然后读取最后一次,需要判断的是:1、如果刚好读完了,直接返回结果

                    2、判断n-result和读取数目num的大小,选择小的,读取并返回数目。

/* 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) {
if (n < 1){
return 0;
}
int time = n / 4;
int result = 0;
char[] chars = new char[4];
for (int i = 0; i < time; i++){
int num = read4(chars);
for (int j = 0; j < num; j++){
buf[i * 4 + j] = chars[j];
}
if (num != 4){
result += num;
return result;
} else {
result += 4;
}
}
if (n - result == 0){
return result;
}
int num = read4(chars);
for (int i = 0; i < Math.min(n - result, num); i++){
buf[result + i] = chars[i];
}
result += Math.min(n - result, num);
return result;
}
}