受Raymond Chen帖子的启发，假设你有一个4x4二维数组，写一个函数使它旋转90度。Raymond链接到伪代码中的解决方案，但我想看到一些真实的东西。
[1][2][3][4]
[5][6][7][8]
[9][0][1][2]
[3][4][5][6]
就变成:
[3][9][5][1]
[4][0][6][2]
[5][1][7][3]
[6][2][8][4]
更新:Nick的答案是最直接的，但是有没有比n²更好的方法呢?如果矩阵是10000x10000呢?
当前回答
在python中:
import numpy as np
a = np.array(
[
[1, 2, 3, 4],
[5, 6, 7, 8],
[9, 0, 1, 2],
[3, 4, 5, 6]
]
)
print(a)
print(b[::-1, :].T)
其他回答
在python中:
import numpy as np
a = np.array(
[
[1, 2, 3, 4],
[5, 6, 7, 8],
[9, 0, 1, 2],
[3, 4, 5, 6]
]
)
print(a)
print(b[::-1, :].T)
PHP:
array_unshift($array, null);
$array = call_user_func_array("array_map", $array);
如果你需要旋转矩形二维阵列90度，在上面的代码之前或之后添加以下一行(取决于你需要的旋转方向):
$array = array_reverse($array);
/* 90-degree clockwise:
temp_array = left_col
left_col = bottom_row
bottom_row = reverse(right_col)
reverse(right_col) = reverse(top_row)
reverse(top_row) = temp_array
*/
void RotateClockwise90(int ** arr, int lo, int hi) {
if (lo >= hi)
return;
for (int i=lo; i<hi; i++) {
int j = lo+hi-i;
int temp = arr[i][lo];
arr[i][lo] = arr[hi][i];
arr[hi][i] = arr[j][hi];
arr[j][hi] = arr[lo][j];
arr[lo][j] = temp;
}
RotateClockwise90(arr, lo+1, hi-1);
}
这个解决方案不关心正方形或矩形的尺寸，你可以旋转4x5或5x4甚至4x4，它也不关心大小。 注意，这种实现在每次调用rotate90方法时都会创建一个新数组，它根本不会改变原始数组。
public static void main(String[] args) {
int[][] a = new int[][] {
{ 1, 2, 3, 4 },
{ 5, 6, 7, 8 },
{ 9, 0, 1, 2 },
{ 3, 4, 5, 6 },
{ 7, 8, 9, 0 }
};
int[][] rotate180 = rotate90(rotate90(a));
print(rotate180);
}
static int[][] rotate90(int[][] a) {
int[][] ret = new int[a[0].length][a.length];
for (int i = 0; i < a.length; i++) {
for (int j = 0; j < a[i].length; j++) {
ret[j][a.length - i - 1] = a[i][j];
}
}
return ret;
}
static void print(int[][] array) {
for (int i = 0; i < array.length; i++) {
System.out.print("[");
for (int j = 0; j < array[i].length; j++) {
System.out.print(array[i][j]);
System.out.print(" ");
}
System.out.println("]");
}
}
c#代码将[n,m] 2D数组向右旋转90度
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
namespace MatrixProject
{
// mattrix class
class Matrix{
private int rows;
private int cols;
private int[,] matrix;
public Matrix(int n){
this.rows = n;
this.cols = n;
this.matrix = new int[this.rows,this.cols];
}
public Matrix(int n,int m){
this.rows = n;
this.cols = m;
this.matrix = new int[this.rows,this.cols];
}
public void Show()
{
for (var i = 0; i < this.rows; i++)
{
for (var j = 0; j < this.cols; j++) {
Console.Write("{0,3}", this.matrix[i, j]);
}
Console.WriteLine();
}
}
public void ReadElements()
{
for (var i = 0; i < this.rows; i++)
for (var j = 0; j < this.cols; j++)
{
Console.Write("element[{0},{1}]=",i,j);
this.matrix[i, j] = Convert.ToInt32(Console.ReadLine());
}
}
// rotate [n,m] 2D array by 90 deg right
public void Rotate90DegRight()
{
// create a mirror of current matrix
int[,] mirror = this.matrix;
// create a new matrix
this.matrix = new int[this.cols, this.rows];
for (int i = 0; i < this.rows; i++)
{
for (int j = 0; j < this.cols; j++)
{
this.matrix[j, this.rows - i - 1] = mirror[i, j];
}
}
// replace cols count with rows count
int tmp = this.rows;
this.rows = this.cols;
this.cols = tmp;
}
}
class Program
{
static void Main(string[] args)
{
Matrix myMatrix = new Matrix(3,4);
Console.WriteLine("Enter matrix elements:");
myMatrix.ReadElements();
Console.WriteLine("Matrix elements are:");
myMatrix.Show();
myMatrix.Rotate90DegRight();
Console.WriteLine("Matrix rotated at 90 deg are:");
myMatrix.Show();
Console.ReadLine();
}
}
}
结果:
Enter matrix elements:
element[0,0]=1
element[0,1]=2
element[0,2]=3
element[0,3]=4
element[1,0]=5
element[1,1]=6
element[1,2]=7
element[1,3]=8
element[2,0]=9
element[2,1]=10
element[2,2]=11
element[2,3]=12
Matrix elements are:
1 2 3 4
5 6 7 8
9 10 11 12
Matrix rotated at 90 deg are:
9 5 1
10 6 2
11 7 3
12 8 4
