日期:2014-05-20  浏览次数:20666 次

有关2为数列的问题,求帮助
要求用一个类确定二位数列中最大值和最大值的位置。这个类包含public data field row,column,and maxValue去储存最大值和它的index,这个数列的row和column是整数类型,maxValue是double类型。使用public static Location locationLargest(double [][] a)作为类函数获得二维函数中最大值的位置。获得的值是一个instance of location. 用一个test 程序让用户输入一个二位数列,别且显示数列中最大值的位置。
以下是sample:
Enter the number of rows and columns of the array: 3 4
enter the array
35 2 10 4.3
23.5 4.5 3 45
33 44 5.5 9.6
The location of the largest element is 45 at (2,4)


哪位大神能帮忙做一下么,谢谢了。

------解决方案--------------------
Java code

import java.util.Scanner;

public class Test {  
    static class Location{
        int row,col;
        double max;
        
        public Location(int row,int col,double max){
            this.row = row;
            this.col = col;
            this.max = max;
        }

        public int getRow() {
            return row;
        }

        public int getCol() {
            return col;
        }

        public double getMax() {
            return max;
        }
    }
    
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        double[][] arr = null;
        try {
            System.out.println("Enter the number of rows and columns of the array:");
            String input = scanner.next();
            int row = Integer.valueOf(input);
            input = scanner.next();
            int col = Integer.valueOf(input);
            arr = new double[row][col];
            System.out.println("enter the array");
            for(int i = 0;i < row;i++){
                for(int j = 0;j < col;j++){
                    input = scanner.next();
                    arr[i][j] = Double.valueOf(input);
                }
            }
        } catch (Exception e) {
            System.out.println("input error");
        }
        Location location = locationLargest(arr);
        System.out.println("The location of the largest element is " + 
                location.getMax() + " at (" + (location.getRow() + 1) + "," + 
                (location.getCol() + 1) + ")");
    }
    
    static Location locationLargest(double[][] a){
        Location location = new Location(0,0,a[0][0]);
        for(int i = 0;i < a.length;i++){
            for(int j = 0;j < a[0].length;j++){
                if(a[i][j] > location.getMax()){
                    location = new Location(i,j,a[i][j]);
                }
            }
        }
        return location;
    }
}

------解决方案--------------------
Java code

import java.util.Scanner;

public class TestSync {

    public static void main(String args[]) {

        Scanner s = new Scanner(System.in);
        int countX = 0, countY = 0;
        double[][] data = null;
        for (;;) {
            System.out
                    .print("Enter the number of rows and columns of the array:");
            String str = s.nextLine();
            try {
                String[] temp = str.split("\\s+");
                data = new double[Integer.valueOf(temp[0])][Integer
                        .valueOf(temp[1])];
            } catch (Exception e) {
                continue;
            }
            break;
        }
        countX = data.length;
        countY = data[0].length;
        double max = Double.MIN_NORMAL;
        int maxIndexX = -1;
        int maxIndexY = -1;
        boolean exitFlag = false;
        while (!exitFlag) {
            System.out.println("Enter the array:");
            exitFlag = true;
            for (int index = 0; index < countX * countY; index++) {
                try {
                    data[index / countY][index % countY] = s.nextDouble();
                    if (max < data[index / countY][index % countY]) {
                        maxIndexX = index / countY;
                        maxIndexY = index % countY;
                        max = data[index / countY][index % countY];
                    }
                } catch (Exception e) {
                    exitFlag = false;
                    break;
                }
            }
        }
        System.out.println("The location of the largest element is " + max
                + " at ( " + maxIndexX + "," + maxIndexY + ")");
    }
}