矩阵操作试题(C++/Python)——矩阵元素逆时针旋转90度

0. 前言

给出一个矩阵，得到他的转置矩阵，输入以及要求输出如下： e.g.0.1 示例1 3*3矩阵

Input
1    2    3
4    5    6
7    8    9

Output:
3 	6	9 
2	5	8 
1	4	7

e.g.0.2 示例2 4*4矩阵

Input:
1    2    3    4    
5    6    7    8
9    10   11   12
13   14   15   16

Output:
4	8	12	16 
3	7	11	15 
2	6	10	14 
1	5	9	13

1. 程序C++版

Code.1.1 示例程序C++版

#include <iostream> 
#define N 4 
using namespace std; 

void displayMatrix(int mat[N][N]); 

void rotateMatrix(int mat[][N]) 
{ 
	for (int x = 0; x < N / 2; x++) 
	{ 
		for (int y = x; y < N-x-1; y++) 
		{ 
			int temp = mat[x][y]; 
			mat[x][y] = mat[y][N-1-x];
			mat[y][N-1-x] = mat[N-1-x][N-1-y]; 
			mat[N-1-x][N-1-y] = mat[N-1-y][x]; 
			mat[N-1-y][x] = temp; 
		} 
	} 
} 

void displayMatrix(int mat[N][N]) 
{ 
	for (int i = 0; i < N; i++) 
	{ 
		for (int j = 0; j < N; j++) 
			cout << mat[i][j]<< " ";
		cout << endl; 
	} 
	cout << endl;  
} 

int main() 
{ 
	int mat[N][N] = 
	{ 
		{1, 2, 3, 4}, 
		{5, 6, 7, 8}, 
		{9, 10, 11, 12}, 
		{13, 14, 15, 16} 
	}; 
	rotateMatrix(mat); 
	displayMatrix(mat); 

	return 0; 
}

1. 程序Python版

Code.1.1 示例程序Python版

N = 4
def rotateMatrix(mat): 
	for x in range(0, int(N/2)): 
		for y in range(x, N-x-1): 
			temp = mat[x][y] 
			mat[x][y] = mat[y][N-1-x] 
			mat[y][N-1-x] = mat[N-1-x][N-1-y] 
			mat[N-1-x][N-1-y] = mat[N-1-y][x] 
			mat[N-1-y][x] = temp 

def displayMatrix( mat ): 
	for i in range(0, N): 
		for j in range(0, N): 
			print (mat[i][j], end = ' ') 
		print ("") 
		
mat = [[0 for x in range(N)] for y in range(N)] 
mat = [ [1, 2, 3, 4 ], 
		[5, 6, 7, 8 ], 
		[9, 10, 11, 12 ], 
		[13, 14, 15, 16 ] ] 

rotateMatrix(mat) 
displayMatrix(mat)

升级版见：矩阵操作试题(C++/Python)——矩阵元素逆时针旋转90度（升级版）