从文本文件读取到数组

Hi im currently trying to do a hackerearth challenge sum of medians and it involves me reading from a text file and storing the values in an array. The first value has to be stored in a variable N which i am able to do but the the remaining values have to be stored in an array. This is where i become stuck. i have to read each value line by line and then store it in the array . this is my code that i have been trying to get it working on but i just cant see where im going wrong.

嗨我目前正在尝试做一个hackerearth挑战中位数的总和，它涉及我从文本文件中读取并将值存储在数组中。第一个值必须存储在我能够做的变量N中，但其余的值必须存储在一个数组中。这就是我陷入困境的地方。我必须逐行读取每个值，然后将其存储在数组中。这是我的代码，我一直试图让它工作，但我只是不知道我哪里出错了。

import java.io.BufferedReader; 
import java.io.InputStreamReader; 

class TestClass { 
 public static void main(String args[] ) throws Exception { 

 // read number of data from system standard input. 
 BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); 
 String line = br.readLine(); 
 int N = Integer.parseInt(line); 
 int i = 1;
 int[] myIntArray = new int[N];
  // median sum 
 long SumMedians = 0; 
 int median = 0;


 while (i<N)

     //read one line file and parse as an integer
     //store the value in an array
 { 

     myIntArray [i] = Integer.parseInt(line);

 i = i + 1; // increment i so i is the total numbers read
 }

so as i said i must increment through the text file storing each value on the line in an array. Any help would be amazing thanks

所以我说我必须通过文本文件增加存储数组中的行上的每个值。任何帮助都会非常棒，谢谢

The text file will look like this

文本文件将如下所示

五

one string per line, which i have to pass into an integer. what i will be doing is after i store the value from the line into the array i will be sorting it and finding its medium and then repeat this process until all the values from the text file have been read.

每行一个字符串，我必须传递给一个整数。我将要做的是在我将行中的值存储到数组中后，我将对其进行排序并找到其介质，然后重复此过程，直到读取了文本文件中的所有值。

The problem which i am trying to do is this one

我想要做的问题就是这个问题

http://www.hackerearth.com/problem/algorithm/sum-of-medians-1/

4 个解决方案

#1

If you're reading from a text file (and not from standard input which is what you're doing at the moment) then you want something like:

如果您正在从文本文件中读取（而不是从您当前正在执行的标准输入中读取），那么您需要以下内容：

// Warning: this could fail if the filename is invaild.
BufferedReader br = new BufferedReader(new FileReader("inputFileName.txt"));

To then read in each line, you can use the following in the while loop:

然后读入每一行，您可以在while循环中使用以下内容：

// Warning: this will crash the program if the line contains anything other than integers.
myIntArray[i] = Integer.parseInt(br.readLine())
 i = i + 1; // increment i so i is the total numbers read

You should also close the reader at the end:

您还应该在最后关闭阅读器：

try{
  br.close();
} catch (IOException e)
{
  System.out.println("Error, program exit!");
  System.exit(1);
}

The import should be swapped from import java.io.InputStreamReader to: import java.io.FileReader

导入应该从import java.io.InputStreamReader交换到：import java.io.FileReader

#2

Since you are only reading 1 line therefore I suspect it to be a single line delimited by colon/semicolon or other character.. try looking into StringTokenizer and Scanner classes

因为你只读了1行所以我怀疑它是由冒号/分号或其他字符分隔的单行。试着查看StringTokenizer和Scanner类

#3

N = the number from parsing a string to a number In the first part of your program it N = 5

N =从解析字符串到数字的数字在程序的第一部分中，N = 5

Why are you using while(i<5)?

你为什么在使用while（i <5）？

If anything you should be

如果你应该做任何事情

r = number of lines in text file;

while (i< r)
{

readline;

parseline;

store in array;

}

and then sort

然后排序

#4

Adapting the example they gave you

调整他们给你的例子

import java.io.BufferedReader;
import java.io.InputStreamReader;

class TestClass {
    public static void main(String args[] ) throws Exception {
        /*
         * Read input from stdin and provide input before running
         */

        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String line = br.readLine();
        int N = Integer.parseInt(line);

        //create storage array
        int[] myIntArray = new int[N];

        //read remainder of file
        for (int i = 0; i < N; i++) {
            String line = br.readLine();
            myIntArray[i] = Integer.parseInt(line);
        }
        // close file
        br.close();


        //Perform median calculations
        int median = 0;
        ...
        System.out.println(median);
}

}

#1