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
在原地顺时针90度旋转使用矢量矢量..
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
//Rotate a Matrix by 90 degrees
void rotateMatrix(vector<vector<int> > &matrix){
int n=matrix.size();
for(int i=0;i<n;i++){
for(int j=i+1;j<n;j++){
swap(matrix[i][j],matrix[j][i]);
}
}
for(int i=0;i<n;i++){
reverse(matrix[i].begin(),matrix[i].end());
}
}
int main(){
int n;
cout<<"enter the size of the matrix:"<<endl;
while (cin >> n) {
vector< vector<int> > m;
cout<<"enter the elements"<<endl;
for (int i = 0; i < n; i++) {
m.push_back(vector<int>(n));
for (int j = 0; j < n; j++)
scanf("%d", &m[i][j]);
}
cout<<"the rotated matrix is:"<<endl;
rotateMatrix(m);
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++)
cout << m[i][j] << ' ';
cout << endl;
}
}
return 0;
}
其他回答
#include <iostream>
#include <iomanip>
using namespace std;
const int SIZE=3;
void print(int a[][SIZE],int);
void rotate(int a[][SIZE],int);
void main()
{
int a[SIZE][SIZE]={{11,22,33},{44,55,66},{77,88,99}};
cout<<"the array befor rotate\n";
print(a,SIZE);
rotate( a,SIZE);
cout<<"the array after rotate\n";
print(a,SIZE);
cout<<endl;
}
void print(int a[][SIZE],int SIZE)
{
int i,j;
for(i=0;i<SIZE;i++)
for(j=0;j<SIZE;j++)
cout<<a[i][j]<<setw(4);
}
void rotate(int a[][SIZE],int SIZE)
{
int temp[3][3],i,j;
for(i=0;i<SIZE;i++)
for(j=0;j<SIZE/2.5;j++)
{
temp[i][j]= a[i][j];
a[i][j]= a[j][SIZE-i-1] ;
a[j][SIZE-i-1] =temp[i][j];
}
}
PHP:
<?php
$a = array(array(1,2,3,4),array(5,6,7,8),array(9,0,1,2),array(3,4,5,6));
$b = array(); //result
while(count($a)>0)
{
$b[count($a[0])-1][] = array_shift($a[0]);
if (count($a[0])==0)
{
array_shift($a);
}
}
从PHP5.6开始，数组转位可以通过一个狡猾的array_map()调用来执行。换句话说，列被转换为行。
代码:(演示)
$array = [
[1, 2, 3, 4],
[5, 6, 7, 8],
[9, 0, 1, 2],
[3, 4, 5, 6]
];
$transposed = array_map(null, ...$array);
美元转置:
[
[1, 5, 9, 3],
[2, 6, 0, 4],
[3, 7, 1, 5],
[4, 8, 2, 6]
]
这是我对矩阵90度旋转的尝试，这是c中的2步解决方案，首先转置矩阵，然后交换cols。
#define ROWS 5
#define COLS 5
void print_matrix_b(int B[][COLS], int rows, int cols)
{
for (int i = 0; i <= rows; i++) {
for (int j = 0; j <=cols; j++) {
printf("%d ", B[i][j]);
}
printf("\n");
}
}
void swap_columns(int B[][COLS], int l, int r, int rows)
{
int tmp;
for (int i = 0; i <= rows; i++) {
tmp = B[i][l];
B[i][l] = B[i][r];
B[i][r] = tmp;
}
}
void matrix_2d_rotation(int B[][COLS], int rows, int cols)
{
int tmp;
// Transpose the matrix first
for (int i = 0; i <= rows; i++) {
for (int j = i; j <=cols; j++) {
tmp = B[i][j];
B[i][j] = B[j][i];
B[j][i] = tmp;
}
}
// Swap the first and last col and continue until
// the middle.
for (int i = 0; i < (cols / 2); i++)
swap_columns(B, i, cols - i, rows);
}
int _tmain(int argc, _TCHAR* argv[])
{
int B[ROWS][COLS] = {
{1, 2, 3, 4, 5},
{6, 7, 8, 9, 10},
{11, 12, 13, 14, 15},
{16, 17, 18, 19, 20},
{21, 22, 23, 24, 25}
};
matrix_2d_rotation(B, ROWS - 1, COLS - 1);
print_matrix_b(B, ROWS - 1, COLS -1);
return 0;
}
这是将数组旋转90度的简单C代码。希望这能有所帮助。
#include <stdio.h>
void main(){
int arr[3][4] = {85, 2, 85, 4,
85, 6, 7, 85,
9, 85, 11, 12};
int arr1[4][3];
int i = 0, j = 0;
for(i=0;i<4;i++){
int k = 2;//k = (number of columns in the new array arr1 - 1)
for(j=0;j<3;j++){
arr1[i][j]=arr[k][i];
k--;
}
}
int l, m;
for(l=0;l<4;l++){
for(m=0;m<3;m++){
printf("%d ", arr1[l][m]);
}
printf("\n");
}
}//end main
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
