将每个 Matrix 元素替换为其行产品，但该元素除外

给定一个M*N阶的二维数组mat[][] 。任务是用同一行其他元素的乘积替换每一行的每个元素。

例子：

Input: mat[][] = {{3, 4, 1}, {6, 1, 7}, {2, 9, 8}}
Output:
4, 3, 12
7, 42, 6
72, 16, 18             

Input: mat[][] = {{13, 4, 5}, {6, 11, 1}}
Output:
20, 65, 52
11, 6, 66

方法：这个想法很简单。逐行遍历矩阵并找到每一行的乘积。在第二次遍历中，找到行中每个单元格的替换值。请按照以下步骤解决问题：

• 初始化变量mul、i和j。
• 使用变量i遍历范围[0, M)并执行以下任务：
  • 将mul的值设置为1。
  • 使用变量j迭代范围[0, N)并执行以下任务：
    • 将 mul 的值设置为mul *mat[i][j]。
  • 使用变量j迭代范围[0, N)并执行以下任务：
    • 将mat[i][j]的值设置为mul/mat[i][j]。
• 执行上述步骤后，打印mat[][]的值作为答案。

下面是上述方法的实现。

C++

// C++ program for the above approach
#include 
using namespace std;
const int M = 3, N = 3;
 
// Show Matrix.
void showMatrix(vector >& mat)
{
    int i, j;
    for (i = 0; i < M; i++) {
        for (j = 0; j < N; j++) {
            cout << mat[i][j] << " ";
        }
        cout << endl;
    }
}
 
// Utility Function
void ReplaceWithProduct(vector >& mat)
{
    int mul, i, j;
    for (i = 0; i < M; i++) {
        mul = 1;
        for (j = 0; j < N; j++)
            mul *= mat[i][j];
        for (j = 0; j < N; j++)
            mat[i][j] = mul / mat[i][j];
    }
    showMatrix(mat);
}
 
// Utility Main function.
int main()
{
    vector > mat = { { 3, 6, 2 },
                                 { 1, 4, 9 },
                                 { 5, 3, 8 } };
    ReplaceWithProduct(mat);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
public class GFG
{
  static int M = 3, N = 3;
 
  // Show Matrix.
  static void showMatrix(int [][]mat)
  {
    int i, j;
    for (i = 0; i < M; i++) {
      for (j = 0; j < N; j++) {
        System.out.print(mat[i][j] + " ");
      }
      System.out.println();
    }
  }
 
  // Utility Function
  static void ReplaceWithProduct(int [][]mat)
  {
    int mul, i, j;
    for (i = 0; i < M; i++) {
      mul = 1;
      for (j = 0; j < N; j++)
        mul *= mat[i][j];
      for (j = 0; j < N; j++)
        mat[i][j] = mul / mat[i][j];
    }
    showMatrix(mat);
  }
 
  // Utility Main function.
  public static void main(String args[])
  {
    int [][]mat = { { 3, 6, 2 },
                   { 1, 4, 9 },
                   { 5, 3, 8 } };
    ReplaceWithProduct(mat);
 
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Python3

# python3 program for the above approach
M = 3
N = 3
 
# Show Matrix.
def showMatrix(mat):
 
    for i in range(0, M):
        for j in range(0, N):
            print(mat[i][j], end=" ")
 
        print()
 
# Utility Function
def ReplaceWithProduct(mat):
 
    for i in range(0, M):
        mul = 1
        for j in range(0, N):
            mul *= mat[i][j]
        for j in range(0, N):
            mat[i][j] = mul // mat[i][j]
 
    showMatrix(mat)
 
# Utility Main function.
if __name__ == "__main__":
 
    mat = [[3, 6, 2],
           [1, 4, 9],
           [5, 3, 8]]
    ReplaceWithProduct(mat)
 
# This code is contributed by rakeshsahni

C#

// C# program for the above approach
using System;
public class GFG
{
  static int M = 3, N = 3;
 
  // Show Matrix.
  static void showMatrix(int [,]mat)
  {
    int i, j;
    for (i = 0; i < M; i++) {
      for (j = 0; j < N; j++) {
        Console.Write(mat[i,j] + " ");
      }
      Console.WriteLine();
    }
  }
 
  // Utility Function
  static void ReplaceWithProduct(int [,]mat)
  {
    int mul, i, j;
    for (i = 0; i < M; i++) {
      mul = 1;
      for (j = 0; j < N; j++)
        mul *= mat[i,j];
      for (j = 0; j < N; j++)
        mat[i,j] = mul / mat[i,j];
    }
    showMatrix(mat);
  }
 
  // Utility Main function.
  public static void Main(String []args)
  {
    int [,]mat = { { 3, 6, 2 },
                   { 1, 4, 9 },
                   { 5, 3, 8 } };
    ReplaceWithProduct(mat);
 
  }
}
 
// This code is contributed by 29AjayKumar

Javascript

      // JavaScript code for the above approach
 
      let M = 3, N = 3;
 
      // Show Matrix.
      function showMatrix(mat) {
          let i, j;
          for (i = 0; i < M; i++) {
              for (j = 0; j < N; j++) {
                  document.write(mat[i][j] + " ");
              }
              document.write('<br>')
          }
      }
 
      // Utility Function
      function ReplaceWithProduct(mat) {
          let mul, i, j;
          for (i = 0; i < M; i++) {
              mul = 1;
              for (j = 0; j < N; j++)
                  mul *= mat[i][j];
              for (j = 0; j < N; j++)
                  mat[i][j] = mul / mat[i][j];
          }
          showMatrix(mat);
      }
 
      // Utility Main function.
 
      let mat = [[3, 6, 2],
      [1, 4, 9],
      [5, 3, 8]];
      ReplaceWithProduct(mat);
 
// This code is contributed by Potta Lokesh
  

输出：

12 6 18 
36 9 4 
24 40 15

时间复杂度： O(N*N)
辅助空间： O(1)