java 求矩阵的秩 java实现矩阵计算

直接上代码吧。

package chapter2;

import java.util.Scanner;

public class Question2 {
    public static void main(String[] args) {
        int m, n;
        Scanner s = new Scanner(System.in);
        System.out.println("请输入矩阵的行数：");
        m = s.nextInt();
        System.out.println("请输入矩阵的列数：");
        n = s.nextInt();
        int[][] a = new int[m][n], b = new int[m][n];
        int[][] result = new int[m][n];
        System.out.println("输入第一个矩阵的数");
        scan(a);
        System.out.println("输入第二个矩阵的数");
        scan(b);
        System.out.println("输入想进行的计算方式（+，-，*）");
        char computer = s.next().charAt(0);  //从键盘输入一个char类型字符的方法
        compute(a, b, result, computer);
        System.out.println("计算结果是：");
        print(result);
    }

    //计算结果的输出
    private static void print(int[][] result) {
        for (int i = 0; i < result.length; i++) {
            for (int j = 0; j < result[i].length; j++) {
                System.out.printf("%d\t", result[i][j]);
            }
            System.out.println("\n");
        }
    }

    //矩阵的计算
    private static void compute(int[][] a, int[][] b, int[][] result, char computer) {
        for (int i = 0; i < result.length; i++) {
            for (int j = 0; j < result[i].length; j++) {
                switch (computer) {
                    case '+':
                        result[i][j] = a[i][j] + b[i][j];
                        break;
                    case '-':
                        result[i][j] = a[i][j] - b[i][j];
                        break;
                    case '*':
                        result[i][j] = a[i][j] * b[i][j];
                        break;
                    default:
                        System.out.println("计算方式错误");
                        break;
                }
            }
        }
    }

    //给两个矩阵赋初值
    private static void scan(int[][] a) {
        Scanner s = new Scanner(System.in);
        for (int i = 0; i < a.length; i++) {
            for (int j = 0; j < a[i].length; j++) {
                a[i][j] = s.nextInt();

            }
        }
    }
}