Java实现矩阵加减乘除及转制等运算功能示例
作者:miangangzhen 时间:2023-07-05 04:49:53
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:
希望本文所述对大家java程序设计有所帮助。
来源:http://blog.csdn.net/miangangzhen/article/details/52275480
标签:Java,矩阵
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