Java实现矩阵加减乘除及转制等运算功能示例

时间:2019-04-12
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本文实例讲述了Java实现矩阵加减乘除及转制等运算功能。分享给大家供大家参考,具体如下:

Java初学,编写矩阵预算程序,当做工具,以便以后写算法时使用。

public class MatrixOperation {
  public static int[][] add(int[][] matrix_a, int[][] matrix_b) {
    int row = matrix_a.length;
    int col = matrix_a[0].length;
    int[][] result = new int[row][col];
    if (row != matrix_b.length || col != matrix_b[0].length) {
      System.out.println("Fault");
    } else {
      for (int i = 0; i < row; i++) {
        for (int j = 0; j < col; j++) {
          result[i][j] = matrix_a[i][j] + matrix_b[i][j];
        }
      }
    }
    return result;
  }
  public static int[][] sub(int[][] matrix_a, int[][] matrix_b) {
    int row = matrix_a.length;
    int col = matrix_a[0].length;
    int[][] result = new int[row][col];
    if (row != matrix_b.length || col != matrix_b[0].length) {
      System.out.println("Fault");
    } else {
      for (int i = 0; i < row; i++) {
        for (int j = 0; j < col; j++) {
          result[i][j] = matrix_a[i][j] - matrix_b[i][j];
        }
      }
    }
    return result;
  }
  public static int[][] dot(int[][] matrix_a, int[][] matrix_b) {
    /*
     * matrix_a's dimention m*p matrix_b's dimention p*n. return dimention
     * m*n
     */
    int row = matrix_a.length;
    int col = matrix_a[0].length;
    int[][] result = new int[row][col];
    if (col != matrix_b.length) {
      System.out.println("Fault");
    } else {
      for (int i = 0; i < row; i++) {
        for (int j = 0; j < col; j++) {
          result[i][j] = 0;
          for (int k = 0; k < col; k++) {
            result[i][j] += matrix_a[i][k] * matrix_b[k][j];
          }
        }
      }
    }
    return result;
  }
  public static int[][] dot(int[][] matrix_a, int b) {
    int row = matrix_a.length;
    int col = matrix_a[0].length;
    int[][] result = new int[row][col];
    for (int i = 0; i < row; i++) {
      for (int j = 0; j < col; j++) {
        result[i][j] = matrix_a[i][j] * b;
      }
    }
    return result;
  }
  public static int[][] mul(int[][] matrix_a, int[][] matrix_b) {
    /*
     * matrix_a's dimention m*n matrix_b's dimention m*n. return dimention
     * m*n
     */
    int row = matrix_a.length;
    int col = matrix_a[0].length;
    int[][] result = new int[row][col];
    if (row != matrix_b.length || col != matrix_b[0].length) {
      System.out.println("Fault");
    } else {
      for (int i = 0; i < row; i++) {
        for (int j = 0; j < col; j++) {
          result[i][j] = matrix_a[i][j] * matrix_b[i][j];
        }
      }
    }
    return result;
  }
  public static int[][] transport(int[][] matrix_a) {
    int row = matrix_a.length;
    int col = matrix_a[0].length;
    int[][] result = new int[row][col];
    for (int i = 0; i < row; i++) {
      for (int j = 0; j < col; j++) {
        result[j][i] = matrix_a[i][j];
      }
    }
    return result;
  }
  public static void print(int[][] matrix) {
    int row = matrix.length;
    int col = matrix[0].length;
    for (int i = 0; i < row; i++) {
      System.out.print("[");
      for (int j = 0; j < col; j++) {
        System.out.print(matrix[i][j]);
        if (j != col - 1) {
          System.out.print(", ");
        }
      }
      System.out.print("]\n");
    }
  }
  public static void main(String[] args) {
    int[][] a = { { 1, 2 }, { 3, 4 } };
    int[][] b = { { 7, 8 }, { 6, 5 } };
    int[][] c = add(a, b);
    System.out.println("脚本之家测试结果如下:");
    System.out.println("matrix a = ");
    print(a);
    System.out.println("matrix b = ");
    print(b);
    System.out.println("matrix a + b = ");
    print(c);
    c = sub(a, b);
    System.out.println("matrix a - b = ");
    print(c);
    int[][] d = dot(a, b);
    System.out.println("matrix a dot b = ");
    print(d);
    int[][] e = dot(a, 3);
    System.out.println("matrix a * 3 = ");
    print(e);
    int[][] f = transport(a);
    System.out.println("matrix a.T = ");
    print(f);
    int[][] g = mul(a, b);
    System.out.println("matrix a * b = ");
    print(g);
  }
}

运行结果:

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希望本文所述对大家java程序设计有所帮助。