两个矩阵相乘的 Java 程序
给定两个矩阵，将它们相乘的任务。矩阵可以是正方形或矩形。
示例:
Input : mat1[][] = {{1, 2},
{3, 4}}
mat2[][] = {{1, 1},
{1, 1}}
Output : {{3, 3},
{7, 7}}
Input : mat1[][] = {{2, 4},
{3, 4}}
mat2[][] = {{1, 2},
{1, 3}}
Output : {{6, 16},
{7, 18}}
方阵乘法: 下面的程序将两个大小为 4*4 的方阵相乘，我们可以针对不同的维度改变 N。
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to multiply two square
// matrices.
import java.io.*;
class GFG {
static int N = 4;
// This function multiplies mat1[][]
// and mat2[][], and stores the result
// in res[][]
static void multiply(int mat1[][],
int mat2[][], int res[][])
{
int i, j, k;
for (i = 0; i < N; i++) {
for (j = 0; j < N; j++) {
res[i][j] = 0;
for (k = 0; k < N; k++)
res[i][j] += mat1[i][k]
* mat2[k][j];
}
}
}
// Driver code
public static void main(String[] args)
{
int mat1[][] = { { 1, 1, 1, 1 },
{ 2, 2, 2, 2 },
{ 3, 3, 3, 3 },
{ 4, 4, 4, 4 } };
int mat2[][] = { { 1, 1, 1, 1 },
{ 2, 2, 2, 2 },
{ 3, 3, 3, 3 },
{ 4, 4, 4, 4 } };
// To store result
int res[][] = new int[N][N];
int i, j;
multiply(mat1, mat2, res);
System.out.println("Result matrix"
+ " is ");
for (i = 0; i < N; i++) {
for (j = 0; j < N; j++)
System.out.print(res[i][j]
+ " ");
System.out.println();
}
}
}
// This code is contributed by anuj_67.
Output
Result matrix is
10 10 10 10
20 20 20 20
30 30 30 30
40 40 40 40
时间复杂度: O(n 3 )。它可以使用斯特拉森矩阵乘法进行优化
辅助空间: O(n 2 )
矩形矩阵的乘法: 我们使用 C 语言中的指针与矩阵相乘。请参考以下帖子作为代码的先决条件。 如何在 C 中传递一个 2D 阵作为参数？
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to multiply two matrices.
public class GFG
{
/**
* to find out matrix multiplication
*
* @param matrix1 First matrix
* @param rows1 Number of rows in matrix 1
* @param cols1 Number of columns in matrix 1
* @param matrix2 Second matrix
* @param rows2 Number of rows in matrix 2
* @param cols2 Number of columns in matrix 2
* @return the result matrix (matrix 1 and matrix 2
* multiplication)
*/
public static int[][] matrixMultiplication(
int[][] matrix1, int rows1, int cols1,
int[][] matrix2, int rows2, int cols2)
throws Exception
{
// Required condition for matrix multiplication
if (cols1 != rows2) {
throw new Exception("Invalid matrix given.");
}
// create a result matrix
int resultMatrix[][] = new int[rows1][cols2];
// Core logic for 2 matrices multiplication
for (int i = 0; i < resultMatrix.length; i++)
{
for (int j = 0;
j < resultMatrix[i].length;
j++)
{
for (int k = 0; k < cols1; k++)
{
resultMatrix[i][j]
+= matrix1[i][k] * matrix2[k][j];
}
}
}
return resultMatrix;
}
// Driver code
public static void main(String[] args) throws Exception
{
// Initial matrix 1 and matrix 2
int matrix1[][] = { { 2, 4 }, { 3, 4 } };
int matrix2[][] = { { 1, 2 }, { 1, 3 } };
// Function call to get a matrix multiplication
int resultMatrix[][] = matrixMultiplication(
matrix1, 2, 2, matrix2, 2, 2);
// Display result matrix
System.out.println("Result Matrix is:");
for (int i = 0; i < resultMatrix.length; i++)
{
for (int j = 0;
j < resultMatrix[i].length;
j++)
{
System.out.print(resultMatrix[i][j] + " ");
}
System.out.println();
}
}
// This code is contributed by darshatandel1998 (Darshan
// Tandel)
}
Output
6 16
7 18
时间复杂度: O(n 3 )。可以使用斯特拉森的矩阵乘法进行优化
辅助空间: O(m1 * n2)
更多详情请参考两矩阵相乘程序整篇文章！
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