簡體   English   中英

Java中的鋸齒數組

[英]Jagged array in java

我的最終輸出應為:

How many rows are in the jagged array? 4
Enter a row, separated by spaces: 9 2 14 5 8
Enter a row, separated by spaces: 3
Enter a row, separated by spaces: 15 23
Enter a row, separated by spaces: 9 8 7 6 5 4 3

After the funky operation, the resulting array is:

9    4   42   20   40
6
45   92
36   40   42   42   40   36   30

但是我不斷出錯:

Exception in thread "main" java.util.NoSuchElementException: No line found
    at ScannerHacked.nextLine(ScannerHacked.java:1525)
    at jagged.main(jagged.java:14)

這是我的代碼:

import java.util.Scanner;

public class jagged {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("How many rows are in the jagged array? ");
        int row = sc.nextInt();
        int[][] jaggedArray = new int[row][];

        for(int i = 0; i < row; i++)
        {   Scanner rows = new Scanner(System.in);
            System.out.print("Enter a row, separated by spaces: ");
            String arraystring = rows.nextLine();
            String []a = arraystring.split(" ");
            jaggedArray[i] = new int[a.length];
        for(int j = 0; j < jaggedArray[i].length; j++)
        {
            int y = Integer.parseInt(a[j]);
            jaggedArray[i][j]  =  y;

        }

        }
        System.out.println("After the funky operation, the resulting array is:");
        for (int i = 0; i < row; i++)
        {
            for (int j = 0; j < jaggedArray[i].length; j++)
            {
                if(jaggedArray[i][j] > 9)
                System.out.print("   "+(jaggedArray[i][j]*(i+j+1)) + "");
                    else
                System.out.print("    "+(jaggedArray[i][j]*(i+j+1)) + "");
            }
            System.out.print("\n");
        }   
    }

}

如我所見,您正在main()方法中使用具有相同源(System.in)兩個Scanner ...

Scanner sc = new Scanner(System.in);
Scanner rows = new Scanner(System.in);

它可能會引發異常,因此您應該先關閉Scannersc.close(); 當要使用另一台具有相同來源的Scanner時。

暫無
暫無

聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.

 
粵ICP備18138465號  © 2020-2024 STACKOOM.COM