给定一个表示矩阵大小的整数N ，任务是构造一个方阵N * N ，它具有从1 到 N 2的元素，使得其对角线和的奇偶性等于整数N的奇偶性。
例子：
Input: N = 4
Output:
1 2 3 4
8 7 6 5
9 10 11 12
16 15 14 13
Explanation:
Sum of diagonal = 32 and 36 and integer N = 4, all the numbers are even that is same parity.
Input: N = 3
Output:
1 2 3
6 5 4
7 8 9
Explanation:
Sum of diagonal = 15 and integer N = 3, all the numbers are odd that is same parity.
方法：这个想法是观察在以替代方式填充矩阵中的元素时， N的奇偶性和对角线的总和是相同的。计数器从 1 开始，然后按递增顺序从0 到 N – 1填充第一行，然后从索引N – 1 到 0填充第二行，依此类推。以这种替代方式继续填充从值1 到 N 2 的每个元素以获得所需的矩阵。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to construct a N * N
// matrix based on the given condition
void createMatrix(int N)
{
// Matrix with given sizM
int M[N][N];
// Counter to insert elements
// from 1 to N * N
int count = 1;
for (int i = 0; i < N; i++) {
// Check if it is first row
// or odd row of the matrix
if (i % 2 == 0) {
for (int j = 0; j < N; j++) {
M[i][j] = count;
count++;
}
}
else
{
// Insert elements from
// right to left
for (int j = N - 1;j >= 0; j--){
M[i][j] = count;
count += 1;
}
}
}
// Print the matrix
for (int i = 0; i < N; i++) {
// Traverse column
for (int j = 0; j < N; j++) {
cout << M[i][j] << " ";
}
cout << endl;
}
}
// Driver Code
int main()
{
// Given size of matrix N
int N = 3;
// Function Call
createMatrix(N);
return 0;
} Java
// Java program for the above approach
class GFG{
// Function to construct a N * N
// matrix based on the given condition
static void createMatrix(int N)
{
// Matrix with given sizM
int M[][] = new int[N][N];
// Counter to insert elements
// from 1 to N * N
int count = 1;
for (int i = 0; i < N; i++)
{
// Check if it is first row
// or odd row of the matrix
if (i % 2 == 0)
{
for (int j = 0; j < N; j++)
{
M[i][j] = count;
count++;
}
}
else
{
// Insert elements from
// right to left
for(int j = N - 1; j >= 0; j--){
M[i][j] = count;
count += 1 ;
}
}
}
// Print the matrix
for (int i = 0; i < N; i++)
{
// Traverse column
for (int j = 0; j < N; j++)
{
System.out.print(M[i][j] + " ");
}
System.out.println();
}
}
// Driver Code
public static void main(String[] args)
{
// Given size of matrix N
int N = 3;
// Function Call
createMatrix(N);
}
}
// This code is contributed by Ritik BansalPython3
# Python3 program for the above approach
# Function to construct a N * N
# matrix based on the given condition
def createMatrix(N):
# Matrix with given size
M = [[0] * N for i in range(N)]
# Counter to insert elements
# from 1 to N * N
count = 1
for i in range(N):
# Check if it is first row
# or odd row of the matrix
if (i % 2 == 0):
for j in range(N):
# Insert elements from
# left to right
M[i][j] = count
count += 1
# Condition if it is second
# row or even row
else:
# Insert elements from
# right to left
for j in range(N - 1, -1, -1):
M[i][j] = count
count += 1
# Print the matrix
for i in range(N):
# Traverse column
for j in range(N):
print(M[i][j], end = " ")
print()
# Driver Code
if __name__ == '__main__':
# Given size of matrix N
N = 3
# Function call
createMatrix(N)
# This code is contributed by mohit kumar 29C#
// C# program for
// the above approach
using System;
class GFG{
// Function to construct a N * N
// matrix based on the given condition
static void createMatrix(int N)
{
// Matrix with given sizM
int[,] M = new int[N, N];
// Counter to insert elements
// from 1 to N * N
int count = 1;
for (int i = 0; i < N; i++)
{
// Check if it is first row
// or odd row of the matrix
if (i % 2 == 0)
{
for (int j = 0; j < N; j++)
{
M[i, j] = count;
count++;
}
}
else
{
// Insert elements from
// right to left
for(int j = N - 1; j >= 0; j--)
{
M[i, j] = count;
count += 1;
}
}
}
// Print the matrix
for (int i = 0; i < N; i++)
{
// Traverse column
for (int j = 0; j < N; j++)
{
Console.Write(M[i, j] + " ");
}
Console.WriteLine();
}
}
// Driver Code
public static void Main()
{
// Given size of matrix N
int N = 3;
// Function Call
createMatrix(N);
}
}
// This code is contributed by ChitranayalJavascript
输出：
1 2 3
6 5 4
7 8 9时间复杂度： O(N*N)
辅助空间： O(1)
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