使用所有 2×2 子矩阵的总和将给定矩阵减小 1

给定一个大小为N*N的方阵M[][] ，任务是使用以下操作将该矩阵缩减为大小为(N-1) * (N-1)的矩阵：

• 从N*N矩阵中取出所有2×2子矩阵，
• 将每个子矩阵的和插入到相应位置的大小为N-1 * N-1的结果矩阵L[ ][ ]中。

例子：

Input: M[][] = {{ 1, 2, 3, 4 },  
                        { 4, 5, 6, 7 },  
                        { 7, 8, 9, 10 },  
                        { 10, 11, 12, 13 }}
Output: L[][] = {{12, 16, 20},  
                          {24, 28, 32},  
                          {36, 40, 44}}
Explanation:

Input: M[][] = {{ 3, 6, 3},  
                        { 4, 7, 6},  
                        { 2, 1, 9}}
Output:  L[][] = {{20, 22},  
                           {14, 23}}

方法：可以通过模拟定义操作的过程来解决任务。请按照以下步骤解决问题：

1. 声明空数组l[]来存储缩减矩阵的元素
2. 声明c = s = 0 ， s 用于获取总和， c用于更改列， p[ ]用于在每次迭代后存储总和。
3. 现在 s 是， s = M[ i ][ c ] + M[ i ][ c + 1 ] + M[ i + 1 ][ c ] + M[ i+1 ][ c+1 ]
4. 它将存储 4 个元素，例如索引 [00, 01, 10, 11] 处的元素，然后是 [01, 02, 11, 12] 类似地从第 2 行和第 3 行开始
5. 将第一个总和存储在数组p[]中并递增c并再次分配s=0 。
6. 重复步骤 3，因此我们将得到数组p [ ] 。
7. 将 p 存储在矩阵l [ ]中并重复步骤 2。

以下是该方法的实现：

C++14

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find the desired matrix
vector > checkMat(vector >& M,
                            int N)
{
 
// Empty array to store 2x2 matrix element
vector > l;
 
for (int i = 0; i < N - 1; i++) {
 
    // Initialization
    int c = 0, s = 0;
    vector p;
 
    for (int j = 0; j < N - 1; j++) {
 
    // Taking sum of 4 elements
    s += M[i][j] + M[i][j+1];
    s += M[i + 1][j] + M[i + 1][j+1];
 
    // Appending sum in array p
    p.push_back(s);
 
    // Assigning s to 0, to take sum
    // of another 4 element
    s = 0;
 
    // Increment c
    c++;
    }
    // Append array p[] to l
    l.push_back(p);
}
return l;
}
 
// Driver code
int main()
{
 
// Original matrix
vector > M = { { 1, 2, 3, 4 },
                            { 4, 5, 6, 7 },
                            { 7, 8, 9, 10 },
                            { 10, 11, 12, 13 } };
 
// Size of matrix
int N = M.size();
 
// Calling function
vector > res = checkMat(M, N);
 
// Printing the result
for (int i = 0; i < res.size(); i++) {
    for (int j = 0; j < res[i].size(); j++) {
    cout << res[i][j] << " ";
    }
    cout << "\n";
}
}
 
// This code is contributed by Taranpreet and improved by prophet1999

Java

// Java code for the above approach
import java.util.*;
 
class GFG
{
 
// Function to find the desired matrix
static ArrayList >
    checkMat(int[][] M, int N)
{
 
    // Empty array to store 2x2 matrix element
    ArrayList > l
    = new ArrayList >(2);
    for (int i = 0; i < N - 1; i++) {
 
    // Initialization
    ArrayList p = new ArrayList();
    int c = 0, s = 0;
 
    for (int j = 0; j < N - 1; j++) {
 
        // Taking sum of 4 elements
        s += M[i][j] + M[i][j+1];
        s += M[i + 1][j] + M[i + 1][j+1];
 
        // Appending sum in array p
        p.add(s);
 
        // Assigning s to 0, to take sum
        // of another 4 element
        s = 0;
 
        // Increment c
        c++;
    }
    // Append array p[] to l
    l.add(p);
    }
    return l;
}
 
public static void main(String[] args)
{
 
    // Original matrix
    int[][] M = { { 1, 2, 3, 4 },
                { 4, 5, 6, 7 },
                { 7, 8, 9, 10 },
                { 10, 11, 12, 13 } };
 
    // Size of matrix
    int N = M.length;
 
    // Calling function
    ArrayList > res = checkMat(M, N);
 
    // Printing the result
    for (int i = 0; i < res.size(); i++) {
    for (int j = 0; j < res.get(i).size(); j++) {
        System.out.print(res.get(i).get(j) + " ");
    }
    System.out.println();
    }
}
}
 
// This code is contributed by Potta Lokesh and improved by prophet1999

Python3

# Python program for the above approach
 
# Function to find the desired matrix
def checkMat(M, N):
 
    # Empty array to store 2x2 matrix element
    l = []
 
    for i in range(N-1):
 
        # Initialization
        c = s = 0
        p = []
 
        for j in range(N-1):
 
            # Taking sum of 4 elements
            s += M[i][j]+M[i][j+1]
            s += M[i + 1][j]+M[i + 1][j+1]
 
            # Appending sum in array p
            p.append(s)
 
            # Assigning s to 0, to take sum
            # of another 4 element
            s = 0
 
            # Increment c
            c += 1
 
        # Append array p[] to l
        l.append(p)
    return l
 
# Driver code
if __name__ == "__main__":
 
    # Original matrix
    M = [[1, 2, 3, 4],
        [4, 5, 6, 7],
        [7, 8, 9, 10],
        [10, 11, 12, 13]]
 
    # Size of matrix
    N = len(M)
 
    # Calling function
    res = checkMat(M, N)
    # Printing the result
    for i in res:
        print(*i)
 
# this code is improved by prophet1999

C#

// C# code for the above approach
using System;
using System.Collections.Generic;
   
public class GFG
{
 
  // Function to find the desired matrix
  static List >
    checkMat(int[,] M, int N)
  {
 
    //  Empty array to store 2x2 matrix element
    List > l
      = new List >(2);
    for (int i = 0; i < N - 1; i++) {
 
      //  Initialization
      List p = new List();
      int c = 0, s = 0;
 
      for (int j = 0; j < N - 1; j++) {
 
        //  Taking sum of 4 elements
        s += M[i,c] + M[i,c + 1];
        s += M[i + 1,c] + M[i + 1,c + 1];
 
        //  Appending sum in array p
        p.Add(s);
 
        //  Assigning s to 0, to take sum
        //  of another 4 element
        s = 0;
 
        //  Increment c
        c++;
      }
      //  Append array p[] to l
      l.Add(p);
    }
    return l;
  }
 
  public static void Main(String[] args)
  {
 
    // Original matrix
    int[,] M = { { 1, 2, 3, 4 },
                 { 4, 5, 6, 7 },
                 { 7, 8, 9, 10 },
                 { 10, 11, 12, 13 } };
 
    //  Size of matrix
    int N = M.GetLength(0);
 
    //  Calling function
    List > res = checkMat(M, N);
 
    //  Printing the result
    for (int i = 0; i < res.Count; i++) {
      for (int j = 0; j < res[i].Count; j++) {
        Console.Write(res[i][j] + " ");
      }
      Console.WriteLine();
    }
  }
}
 
// This code is contributed by 29AjayKumar and improved by prophet1999

Javascript

    // JavaScript program for the above approach
 
    // Function to find the desired matrix
    const checkMat = (M, N) => {
 
        // Empty array to store 2x2 matrix element
        let l = [];
 
        for (let i = 0; i < N - 1; ++i) {
 
            // Initialization
            let c = s = 0;
            let p = [];
 
            for (let j = 0; j < N - 1; ++j) {
 
                // Taking sum of 4 elements
                s += M[i][j] + M[i][j+1];
                s += M[i + 1][j] + M[i + 1][j+1];
 
                // Appending sum in array p
                p.push(s);
 
                // Assigning s to 0, to take sum
                // of another 4 element
                s = 0;
 
                // Increment c
                c += 1;
            }
            // document.write(p);
            // document.write("<br/>")
 
            // Append array p[] to l
            l.push(p);
        }
        return l;
    }
 
    // Driver code
 
    // Original matrix
    let M = [[1, 2, 3, 4],
    [4, 5, 6, 7],
    [7, 8, 9, 10],
    [10, 11, 12, 13]];
 
    // Size of matrix
    let N = M.length;
 
    // Calling function
    let res = checkMat(M, N);
    // Printing the result
    for (let i in res) {
        for (let j in res[i]) {
            document.write(`${res[i][j]} `);
        }
        document.write("<br/>");
    }
 
// This code is contributed by rakeshsahni and improved by prophet1999
 

输出

12 16 20
24 28 32
36 40 44

时间复杂度： O(N*N)
辅助空间： O(N*N)