基于给定的模式，使用给定数字 N 的数字构造一个正方形矩阵
原文:https://www . geesforgeks . org/construct-a-square-matrix-使用基于给定模式的给定数字 n 位数/
给定一个整数 N ，任务是构造一个矩阵mat【】【】的大小M x M(“M”是给定整数中的位数)使得矩阵的每个对角线 l 包含相同的位数，根据给定整数中位数的位置放置，然后从后面再次重复上述步骤。
*示例:*
*输入:* N = 123 输出: {{1，2，3}， {2，3，2}， {3，2，1}} 说明:所需矩阵的大小必须为 3*3。N 的位数是 1、2 和 3。沿着对角线放置 1、2 和 3，从左上角单元格到第 n 个对角线，在第 n 个对角线之后放置 2、1，直到最下面的单元格。
*输入:* N = 3219 输出: {{3，2，1，9}，{2，1，9，1}，{1，9，1，2}，{9，1，2，3}
*方法:*该任务可以通过以对角线方式遍历矩阵并根据给定数字中的相应数字分配单元值来解决。
- 提取给定整数的数字并存储在一个向量中，比如 v 。
- 对于矩阵的后半对角线，再次以相反的顺序存储数字。
- 按照所需的顺序分配数字。
- 打印矩阵。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to construct the matrix
void constructMatrix(int n)
{
// Vector to store the
// digits of the integer
vector<int> 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 语言(一种计算机语言，尤用于创建网站)
// 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<Integer> v = new ArrayList<Integer>();
// 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.
Python 3
# 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<int> v = new List<int>();
// 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.
java 描述语言
<script>
// 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
</script>
**Output:
3 2 1 9
2 1 9 1
1 9 1 2
9 1 2 3**
*时间复杂度:O(N2) 辅助空间: O(1)***
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