两个矩阵相乘的 Javascript 程序

原文:https://www . geeksforgeeks . org/JavaScript-程序乘二矩阵/

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 描述语言

<script>

// Javascript program to multiply
// two square matrices.

const N = 4;

// This function multiplies
// mat1[][] and mat2[][], and
// stores the result in res[][]
function multiply(mat1, mat2, res)
{
    let 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
    let i, j;

    // To store result
    let res = new Array(N);
    for (let k = 0; k < N; k++)
        res[k] = new Array(N);

    let mat1 = [ [ 1, 1, 1, 1 ],
                       [ 2, 2, 2, 2 ],
                       [ 3, 3, 3, 3 ],
                       [ 4, 4, 4, 4 ] ];

    let mat2 = [ [ 1, 1, 1, 1 ],
                       [ 2, 2, 2, 2 ],
                       [ 3, 3, 3, 3 ],
                       [ 4, 4, 4, 4 ] ];

    multiply(mat1, mat2, res);

    document.write("Result matrix is <br>");
    for (i = 0; i < N; i++) {
        for (j = 0; j < N; j++)
            document.write(res[i][j] + " ");
        document.write("<br>");
    }

</script>

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 描述语言

<script>

// Javascript program to multiply two
// rectangular matrices

// Multiplies two matrices mat1[][]
// and mat2[][] and prints result.
// (m1) x (m2) and (n1) x (n2) are
// dimensions of given matrices.
function multiply(m1, m2, mat1, n1, n2, mat2)
{
    let x, i, j;
    let res = new Array(m1);
    for (i = 0; i < m1; i++)
        res[i] = new Array(n2);

    for (i = 0; i < m1; i++) 
    {
        for (j = 0; j < n2; j++) 
        {
            res[i][j] = 0;
            for (x = 0; x < m2; x++) 
            {
                res[i][j] += mat1[i][x] * mat2[x][j];
            }
        }
    }
    for (i = 0; i < m1; i++) 
    {
        for (j = 0; j < n2; j++) 
        {
            document.write(res[i][j] + " ");
        }
        document.write("<br>");
    }
}

// Driver code
    let mat1 = [ [ 2, 4 ], [ 3, 4 ] ];
    let mat2 = [ [ 1, 2 ], [ 1, 3 ] ];
    let m1 = 2, m2 = 2, n1 = 2, n2 = 2;

    // Function call
    multiply(m1, m2, mat1, n1, n2, mat2);

</script>

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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