基于给定模式使用给定数字 N 的数字构造方阵

给定一个整数N ，任务是构造一个大小为M x M的矩阵mat[][] （'M' 是给定整数中的位数），使得矩阵的每个对角线l 包含相同的数字，放置根据给定整数中数字的位置，然后再次从后面重复这些步骤。

例子：

Input: N = 123
Output: {{1, 2, 3}, 
               {2, 3, 2}, 
               {3, 2, 1}}
Explanation: The desired matrix must be of size 3*3. The digits of N are 1, 2, and 3. Placing 1, 2 and 3 along the diagonals from the top left cell till the Nth diagonal, and 2, 1 just after the Nth diagonal till the bottom-most cell.

Input: N = 3219
Output: {{3, 2, 1, 9}, {2, 1, 9, 1}, {1, 9, 1, 2}, {9, 1, 2, 3}}

方法：可以通过以对角方式遍历矩阵并根据给定数字中的相应数字分配单元格值来解决该任务。

1. 提取给定整数的数字并将其存储在向量中，例如v 。
2. 再次以相反的顺序存储矩阵的第二半对角线的数字。
3. 按所需顺序分配数字。
4. 打印矩阵。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to construct the matrix
void constructMatrix(int n)
{
    // Vector to store the
    // digits of the integer
    vector v;
 
    // Extracting the digits
    // from the integer
    while (n > 0) {
        v.push_back(n % 10);
        n = n / 10;
    }
 
    // Reverse the vector
    reverse(v.begin(), v.end());
 
    // Size of the vector
    int N = v.size();
 
    // Loop to store the digits in
    // reverse order in the same vector
    for (int i = N - 2; i >= 0; i--) {
        v.push_back(v[i]);
    }
 
    // Matrix to be constructed
    int mat[N][N];
 
    // Assign the digits and
    // print the desired matrix
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < N; j++) {
            mat[i][j] = v[i + j];
            cout << mat[i][j] << " ";
        }
        cout << endl;
    }
}
 
// Driver Code
int main()
{
    int n = 3219;
 
    // Passing n to constructMatrix function
    constructMatrix(n);
 
    return 0;
}

Java

// Java program for the above approach
 
import java.util.ArrayList;
import java.util.Collections;
 
class GFG {
 
    // Function to construct the matrix
    public static void constructMatrix(int n)
    {
       
        // Vector to store the
        // digits of the integer
        ArrayList v = new ArrayList();
 
        // Extracting the digits
        // from the integer
        while (n > 0) {
            v.add(n % 10);
            n = n / 10;
        }
 
        // Reverse the vector
        Collections.reverse(v);
 
        // Size of the vector
        int N = v.size();
 
        // Loop to store the digits in
        // reverse order in the same vector
        for (int i = N - 2; i >= 0; i--) {
            v.add(v.get(i));
        }
 
        // Matrix to be constructed
        int[][] mat = new int[N][N];
 
        // Assign the digits and
        // print the desired matrix
        for (int i = 0; i < N; i++) {
            for (int j = 0; j < N; j++) {
                mat[i][j] = v.get(i + j);
                System.out.print(mat[i][j] + " ");
            }
            System.out.println("");
        }
    }
 
    // Driver Code
    public static void main(String args[]) {
        int n = 3219;
 
        // Passing n to constructMatrix function
        constructMatrix(n);
    }
}
 
// This code is contributed by gfgking.

Python3

# python program for the above approach
 
# Function to construct the matrix
def constructMatrix(n):
 
    # Vector to store the
    # digits of the integer
    v = []
 
    # Extracting the digits
    # from the integer
    while (n > 0):
        v.append(n % 10)
        n = n // 10
 
    # Reverse the vector
    v.reverse()
 
    # Size of the vector
    N = len(v)
 
    # Loop to store the digits in
    # reverse order in the same vector
    for i in range(N-2, -1, -1):
        v.append(v[i])
 
    # Matrix to be constructed
    mat = [[0 for _ in range(N)] for _ in range(N)]
 
    # Assign the digits and
    # print the desired matrix
    for i in range(0, N):
        for j in range(0, N):
            mat[i][j] = v[i + j]
            print(mat[i][j], end=" ")
 
        print()
 
# Driver Code
if __name__ == "__main__":
    n = 3219
 
    # Passing n to constructMatrix function
    constructMatrix(n)
 
# This code is contributed by rakeshsahni

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to construct the matrix
static void constructMatrix(int n)
{
   
    // Vector to store the
    // digits of the integer
    List v = new List();
 
    // Extracting the digits
    // from the integer
    while (n > 0) {
        v.Add(n % 10);
        n = n / 10;
    }
 
    // Reverse the vector
    v.Reverse();
 
    // Size of the vector
    int N = v.Count;
 
    // Loop to store the digits in
    // reverse order in the same vector
    for (int i = N - 2; i >= 0; i--) {
        v.Add(v[i]);
    }
 
    // Matrix to be constructed
    int[,] mat = new int[N, N];
 
    // Assign the digits and
    // print the desired matrix
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < N; j++) {
            mat[i, j] = v[i + j];
            Console.Write(mat[i, j] + " ");
        }
        Console.WriteLine();
    }
}
 
 
// Driver Code
public static void Main()
{
    int n = 3219;
 
    // Passing n to constructMatrix function
    constructMatrix(n);
}
}
 
// This code is contributed by sanjoy_62.

Javascript

       // JavaScript Program to implement
       // the above approach
 
       // Function to construct the matrix
       function constructMatrix(n)
       {
        
           // Vector to store the
           // digits of the integer
           let v = [];
 
           // Extracting the digits
           // from the integer
           while (n > 0) {
               v.push(n % 10);
               n = Math.floor(n / 10);
           }
 
           // Reverse the vector
           v.reverse();
 
           // Size of the vector
           let N = v.length;
 
           // Loop to store the digits in
           // reverse order in the same vector
           for (let i = N - 2; i >= 0; i--) {
               v.push(v[i]);
           }
 
           // Matrix to be constructed
           let mat = new Array(N);
           for (let i = 0; i < mat.length; i++) {
               mat[i] = new Array(N).fill(0);
           }
 
           // Assign the digits and
           // print the desired matrix
           for (let i = 0; i < N; i++) {
               for (let j = 0; j < N; j++) {
                   mat[i][j] = v[i + j];
                   document.write(mat[i][j] + " ");
               }
               document.write("<br>")
           }
       }
 
       // Driver Code
       let n = 3219;
 
       // Passing n to constructMatrix function
       constructMatrix(n);
 
   // This code is contributed by Potta Lokesh
   

输出：

3 2 1 9 
2 1 9 1 
1 9 1 2 
9 1 2 3

时间复杂度： O(N ² )
辅助空间： O(1)