构造一个矩阵，使每个像元由给定矩阵中各个像元的相邻元素之和组成

给定尺寸为N * M的矩阵arr [] [] ，任务是生成一个矩阵，以使任何像元(r，c)存储在给定矩阵中水平，垂直和对角线方向上存在的相邻元素的总和。

例子：

Input: arr[][] = {{1, 3}, {2, 4}}
Output: {{9, 7}, {8, 6}}
Explanation: Matrix is constructed by the following operations: 
For cell (0, 0), arr[1][0] + arr[0][1] + arr[1][1] = 2 + 3 + 4 = 9. 
For cell (0, 1), arr[1][0] + arr[0][0] + arr[1][1] = 2 + 1 + 4 = 7. 
For cell (1, 0), arr[0][0] + arr[0][1] + arr[1][1] = 1 + 3 + 4 = 8. 
For cell (1, 1), arr[1][0] + arr[0][1] + arr[0][0] = 2 + 3 + 1 = 6.

Input: arr[][] = {{1}}
Output: {{0}}

方法：想法是遍历给定矩阵的每个像元，对于每个像元(r，c) ，存储相邻像元{{r-1，c-1}，{r + 1，c + 1}， {r-1，c + 1}，{r + 1，c-1}，{r，c-1}，{r-1，c}，{r + 1，c}，{r，c + 1 }}(如果可能)。

1. 初始化尺寸为N * M的矩阵v [] []以存储每个单元格的结果。
2. 现在，遍历矩阵的每个单元格。对于每个单元格，检查有效的相邻单元格，并继续更新其总和。
3. 遍历后，打印存储在矩阵v [] []的每个单元格中的值。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Initialize rows and columns
int r, c;
 
// Store all 8 directions
vector > dir
    = { { 1, 0 },   { -1, 0 }, { 0, 1 }, { 0, -1 },
        { -1, -1 }, { -1, 1 }, { 1, 1 }, { 1, -1 } };
 
// Function to check if a cell
// (i, j) is valid or not
bool valid(int i, int j)
{
    if (i >= 0 && j >= 0 && i < r && j < c)
        return 1;
 
    return 0;
}
 
// Function to find sum of adjacent cells
// for cell (i, j)
int find(int i, int j, vector >& v)
{
    // Initialize sum
    int s = 0;
 
    // Visit all 8 directions
    for (auto x : dir) {
        int ni = i + x[0], nj = j + x[1];
 
        // Check if cell is valid
        if (valid(ni, nj))
            s += v[ni][nj];
    }
 
    // Return sum
    return s;
}
 
// Function to print sum of adjacent elements
void findsumofneighbors(vector >& M)
{
    // Stores the resultant matrix
    vector > v(r, vector(c, 0));
 
    // Iterate each elements of matrix
    for (int i = 0; i < r; i++) {
        for (int j = 0; j < c; j++) {
 
            // Find adjacent sum
            v[i][j] = find(i, j, M);
            cout << v[i][j] << " ";
        }
        cout << "\n";
    }
}
 
// Driver Code
int main()
{
 
    // Given matrix
    vector > M
        = { { 1, 4, 1 }, { 2, 4, 5 }, { 3, 1, 2 } };
 
    // Size of matrix
    r = M.size(), c = M[0].size();
 
    // Function call
    findsumofneighbors(M);
}

Java

// Java program for the above approach
import java.util.*;
import java.lang.*;
 
public class GFG {
 
    // Initialize rows and columns
    private static int r, c;
 
    // Store all 8 directions
    static int[][] dir
        = { { 1, 0 },   { -1, 0 }, { 0, 1 }, { 0, -1 },
            { -1, -1 }, { -1, 1 }, { 1, 1 }, { 1, -1 } };
 
    // Function to check if a cell
    // (i, j) is valid or not
    public static boolean valid(int i, int j)
    {
        if (i >= 0 && j >= 0 && i < r && j < c)
            return true;
 
        return false;
    }
 
    // Function to find sum of adjacent cells
    // for cell (i, j)
    static int find(int i, int j, int[][] v)
    {
        // Initialize sum
        int s = 0;
 
        // Visit all 8 directions
        for (int k = 0; k < 8; k++) {
 
            int ni = i + dir[k][0], nj = j + dir[k][1];
 
            // Check if cell is valid
            if (valid(ni, nj))
                s += v[ni][nj];
        }
 
        // Return sum
        return s;
    }
 
    // Function to print sum of adjacent elements
    static void findsumofneighbors(int[][] M)
    {
        // Stores the resultant matrix
        int[][] v = new int[r];
 
        // Iterate each elements of matrix
        for (int i = 0; i < r; i++) {
            for (int j = 0; j < c; j++) {
 
                // Find adjacent sum
                v[i][j] = find(i, j, M);
                System.out.print(v[i][j] + " ");
            }
            System.out.println("");
        }
    }
 
    // Driver code
    public static void main(String[] args)
    {
        // Given matrix
        int[][] M
            = { { 1, 4, 1 },
               { 2, 4, 5 },
               { 3, 1, 2 } };
 
        // Size of matrix
        r = M.length;
        c = M[0].length;
 
        // Function call
        findsumofneighbors(M);
    }
}
 
// This code is contributed by ajaykr00kj

Python3

# Python program for the above approach
 
# Initialize rows and columns
r, c = 0, 0;
 
# Store all 8 directions
dir = [[1, 0], [-1, 0], [0, 1], [0, -1],
       [-1, -1], [-1, 1], [1, 1], [1, -1]];
 
# Function to check if a cell
# (i, j) is valid or not
def valid(i, j):
    if (i >= 0 and j >= 0 and i < r and j < c):
        return True;
 
    return False;
 
# Function to find sum of adjacent cells
# for cell (i, j)
def find(i, j, v):
   
    # Initialize sum
    s = 0;
 
    # Visit all 8 directions
    for k in range(8):
 
        ni = i + dir[k][0];
        nj = j + dir[k][1];
 
        # Check if cell is valid
        if (valid(ni, nj)):
            s += v[ni][nj];
 
    # Return sum
    return s;
 
# Function to prsum of adjacent elements
def findsumofneighbors(M):
   
    # Stores the resultant matrix
    v = [[0 for i in range(c)] for j in range(r)];
 
    # Iterate each elements of matrix
    for i in range(r):
        for j in range(c):
           
            # Find adjacent sum
            v[i][j] = find(i, j, M);
            print(v[i][j], end=" ");
 
        print("");
 
# Driver code
if __name__ == '__main__':
   
    # Given matrix
    M = [[1, 4, 1], [2, 4, 5], [3, 1, 2]];
 
    # Size of matrix
    r = len(M[0]);
    c = len(M[1]);
 
    # Function call
    findsumofneighbors(M);
 
# This code is contributed by 29AjayKumar

C#

// C# program for the above approach
using System;
public class GFG {
 
    // Initialize rows and columns
    private static int r, c;
 
    // Store all 8 directions
    static int[,] dir
        = { { 1, 0 },   { -1, 0 }, { 0, 1 }, { 0, -1 },
            { -1, -1 }, { -1, 1 }, { 1, 1 }, { 1, -1 } };
 
    // Function to check if a cell
    // (i, j) is valid or not
    public static bool valid(int i, int j)
    {
        if (i >= 0 && j >= 0 && i < r && j < c)
            return true;
 
        return false;
    }
 
    // Function to find sum of adjacent cells
    // for cell (i, j)
    static int find(int i, int j, int[,] v)
    {
        // Initialize sum
        int s = 0;
 
        // Visit all 8 directions
        for (int k = 0; k < 8; k++) {
 
            int ni = i + dir[k, 0], nj = j + dir[k, 1];
 
            // Check if cell is valid
            if (valid(ni, nj))
                s += v[ni, nj];
        }
 
        // Return sum
        return s;
    }
 
    // Function to print sum of adjacent elements
    static void findsumofneighbors(int[,] M)
    {
        // Stores the resultant matrix
        int[,] v = new int[r, c];
 
        // Iterate each elements of matrix
        for (int i = 0; i < r; i++) {
            for (int j = 0; j < c; j++) {
 
                // Find adjacent sum
                v[i,j] = find(i, j, M);
                Console.Write(v[i, j] + " ");
            }
            Console.WriteLine("");
        }
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        // Given matrix
        int[,] M
            = { { 1, 4, 1 },
               { 2, 4, 5 },
               { 3, 1, 2 } };
 
        // Size of matrix
        r = M.GetLength(0);
        c = M.GetLength(1);
 
        // Function call
        findsumofneighbors(M);
    }
}
 
// This code is contributed by Rajput-Ji

输出

10 13 13 
13 19 12 
7 16 10

时间复杂度： O(N * M)其中N * M是矩阵的维数。
辅助空间： O(N * M)