计算两个矩阵相乘

//时间复杂度O(xyz) 如果是方阵就是O(N^3)
#include <iostream>
using namespace std;
int ** matrix(int** marXY, int** marYZ,int x,int y,int z);
//数据结构作业，计算两个矩阵

int main(){
    //开辟两个矩阵(指针型)
    int** marXY = new int*[2];
    int** marYZ = new int*[2];
    for (int i = 0; i<2; i++) {
        marXY[i] = new int[2];
    }
    for (int i = 0; i<3; i++) {
        marYZ[i] = new int[3];
    }
    //因为矩阵相乘函数是传的指针数组，这里吧
    int marXYcopy[2][2] = {     {2,5},
                                {1,3}

    };
    int marYZcopy[2][3] = {
                                {4,2,3},
                                {1,3,1}

    };
    //把数值型矩阵 传给指针型矩阵
    for (int i = 0; i<2; i++) {
        for (int j = 0; j<2; j++) {
            marXY[i][j] = marXYcopy[i][j];
        }
    }
    for (int i = 0; i<2; i++) {
        for (int j = 0; j<3; j++) {
            marYZ[i][j] = marYZcopy[i][j];
        }
    }
    int **rul =   matrix(marXY, marYZ,2, 2, 3 );
    for (int i = 0; i<2; i++) {
        for (int j = 0; j<3; j++) {
            cout <<rul[i][j]<<" ";
        }
        cout << endl;
    }
    delete rul;
    return 0;
}


//marXY  矩阵x行y列
//marYZ 矩阵y行z列
int ** matrix(int** marXY, int** marYZ,int x,int y,int z){
    //声明个二维数组
    int** newMatrix = new int*[x];
    for (int i = 0; i < y; i++) {
        newMatrix[i] = new int[z];
    }
    //外层newMatrix的x坐标。
    for (int i = 0 ; i < x; i++) {
        //外层newMatrix的y坐标。
        for (int j = 0; j < z; j++) {
            int cur = 0;
            for (int k = 0; k < y; k++) {
                //将第i行 与 第 j列，按照内标的大小，依次相乘后相加。
                cur += marXY[i][k]*marYZ[k][j];
            }
            //新数组的第ij位置的数
            newMatrix[i][j]  = cur;
        }
    }
    return newMatrix;
}