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
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
其他回答
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
已经有很多答案了，我发现两个声称O(1)时间复杂度。真正的O(1)算法是保持数组存储不变，并改变索引其元素的方式。这里的目标是不消耗额外的内存，也不需要额外的时间来迭代数据。
旋转90度，-90度和180度是简单的转换，只要你知道你的2D数组中有多少行和列就可以执行;要将任何向量旋转90度，交换轴并与Y轴相反。对于-90度，交换轴和X轴。对于180度，两个坐标轴都是负的，不交换。
进一步的转换是可能的，例如通过独立地否定轴来水平和/或垂直地镜像。
这可以通过访问器方法来实现。下面的例子是JavaScript函数，但是这些概念同样适用于所有语言。
//按列/行顺序获取数组元素 var getArray2d =函数(a, x, y) { 返回一个[y] [x]; }; / /演示 Var arr = [ [5,4,6]， [1,7,9]， [- 2,11,0]， [8,21， -3]， [3， -1, 2] ]; Var newar = []; arr[0]. foreach (() => newarr。push(新数组(arr.length))); For (var I = 0;I < newar .length;我+ +){ For (var j = 0;J < newarr[0].length;j + +) { newarr[i][j] = getArray2d(arr, i, j); ｝ ｝ console.log (newarr);
// Get an array element rotated 90 degrees clockwise function getArray2dCW(a, x, y) { var t = x; x = y; y = a.length - t - 1; return a[y][x]; } //demo var arr = [ [5, 4, 6], [1, 7, 9], [-2, 11, 0], [8, 21, -3], [3, -1, 2] ]; var newarr = []; arr[0].forEach(() => newarr.push(new Array(arr.length))); for (var i = 0; i < newarr[0].length; i++) { for (var j = 0; j < newarr.length; j++) { newarr[j][i] = getArray2dCW(arr, i, j); } } console.log(newarr);
// Get an array element rotated 90 degrees counter-clockwise function getArray2dCCW(a, x, y) { var t = x; x = a[0].length - y - 1; y = t; return a[y][x]; } //demo var arr = [ [5, 4, 6], [1, 7, 9], [-2, 11, 0], [8, 21, -3], [3, -1, 2] ]; var newarr = []; arr[0].forEach(() => newarr.push(new Array(arr.length))); for (var i = 0; i < newarr[0].length; i++) { for (var j = 0; j < newarr.length; j++) { newarr[j][i] = getArray2dCCW(arr, i, j); } } console.log(newarr);
// Get an array element rotated 180 degrees function getArray2d180(a, x, y) { x = a[0].length - x - 1; y = a.length - y - 1; return a[y][x]; } //demo var arr = [ [5, 4, 6], [1, 7, 9], [-2, 11, 0], [8, 21, -3], [3, -1, 2] ]; var newarr = []; arr.forEach(() => newarr.push(new Array(arr[0].length))); for (var i = 0; i < newarr[0].length; i++) { for (var j = 0; j < newarr.length; j++) { newarr[j][i] = getArray2d180(arr, i, j); } } console.log(newarr);
这段代码假设有一个嵌套数组的数组，其中每个内部数组都是一行。
该方法允许您读取(或写入)元素(甚至是随机顺序)，就像数组已经旋转或转换一样。现在只要选择正确的函数来调用，可能是通过引用，然后就可以了!
这个概念可以扩展为通过访问器方法附加地(非破坏性地)应用转换。包括任意角度旋转和缩放。
在Eigen (c++)中:
Eigen::Matrix2d mat;
mat << 1, 2,
3, 4;
std::cout << mat << "\n\n";
Eigen::Matrix2d r_plus_90 = mat.transpose().rowwise().reverse();
std::cout << r_plus_90 << "\n\n";
Eigen::Matrix2d r_minus_90 = mat.transpose().colwise().reverse();
std::cout << r_minus_90 << "\n\n";
Eigen::Matrix2d r_180 = mat.colwise().reverse().rowwise().reverse(); // +180 same as -180
std::cout << r_180 << "\n\n";
输出:
1 2
3 4
3 1
4 2
2 4
1 3
4 3
2 1
ruby方式:.transpose。地图&:反向
下面是一个原地旋转的数组，而不是使用一个全新的数组来保存结果。我已经停止了数组的初始化和输出。这只适用于正方形数组，但它们可以是任何大小。内存开销等于数组中一个元素的大小，因此您可以对任意大的数组进行旋转。
int a[4][4];
int n = 4;
int tmp;
for (int i = 0; i < n / 2; i++)
{
for (int j = i; j < n - i - 1; j++)
{
tmp = a[i][j];
a[i][j] = a[j][n-i-1];
a[j][n-i-1] = a[n-i-1][n-j-1];
a[n-i-1][n-j-1] = a[n-j-1][i];
a[n-j-1][i] = tmp;
}
}
