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Java中

public class Matrix {
/* Author Shrikant Dande */
private static void showMatrix(int[][] arr,int rows,int col){

    for(int i =0 ;i<rows;i++){
        for(int j =0 ;j<col;j++){
            System.out.print(arr[i][j]+" ");
        }
        System.out.println();
    }

}

private static void rotateMatrix(int[][] arr,int rows,int col){

    int[][] tempArr = new int[4][4];
    for(int i =0 ;i<rows;i++){
        for(int j =0 ;j<col;j++){
            tempArr[i][j] = arr[rows-1-j][i];
            System.out.print(tempArr[i][j]+" ");
        }
        System.out.println();
    }

}
public static void main(String[] args) {
    int[][] arr = { {1,  2,  3,  4},
             {5,  6,  7,  8},
             {9,  1, 2, 5},
             {7, 4, 8, 9}};
    int rows = 4,col = 4;

    showMatrix(arr, rows, col);
    System.out.println("------------------------------------------------");
    rotateMatrix(arr, rows, col);

}

}

其他回答

我的c#示例代码的伟大算法发送@dimple:

/* Author: Dudi,
 * http://www.tutorialspoint.com/compile_csharp_online.php?PID=0Bw_CjBb95KQMYm5qU3VjVGNuZFU */

using System.IO;
using System;

class Program
{
    static void Main()
    {
        Console.WriteLine("Rotating this matrix by 90+ degree:");

        int[,] values=new int[3,3]{{1,2,3}, {4,5,6}, {7,8,9}};
        //int[,] values=new int[4,4]{{101,102,103, 104}, {105,106, 107,108}, {109, 110, 111, 112}, {113, 114, 115, 116}};

        print2dArray(ref values);
        transpose2dArray(ref values);
        //print2dArray(ref values);
        reverse2dArray(ref values);
        Console.WriteLine("Output:");
        print2dArray(ref values);
    }

    static void print2dArray(ref int[,] matrix){
        int  nLen = matrix.GetLength(0);
        int  mLen = matrix.GetLength(1);    
        for(int n=0; n<nLen; n++){
            for(int m=0; m<mLen; m++){
                Console.Write(matrix[n,m] +"\t");
            }
            Console.WriteLine();        
        }
        Console.WriteLine();
    }

    static void transpose2dArray(ref int[,] matrix){
        int  nLen = matrix.GetLength(0);
        int  mLen = matrix.GetLength(1);    
        for(int n=0; n<nLen; n++){
            for(int m=0; m<mLen; m++){
                if(n>m){
                    int tmp = matrix[n,m];
                    matrix[n,m] = matrix[m,n];
                    matrix[m,n] = tmp;
                }
            }
        }
    }

    static void reverse2dArray(ref int[,] matrix){
        int  nLen = matrix.GetLength(0);
        int  mLen = matrix.GetLength(1);
        for(int n=0; n<nLen; n++){
            for(int m=0; m<mLen/2; m++){                
                int tmp = matrix[n,m];
                matrix[n,m] = matrix[n, mLen-1-m];
                matrix[n,mLen-1-m] = tmp;
            }
        }
    }
}

/*
Rotating this matrix by 90+ degree:                                                                                                                                             
1       2       3                                                                                                                                                               
4       5       6                                                                                                                                                               
7       8       9                                                                                                                                                               

Output:                                                                                                                                                                         
7       4       1                                                                                                                                                               
8       5       2                                                                                                                                                               
9       6       3  
*/
#!/usr/bin/env python

original = [ [1,2,3],
             [4,5,6],
             [7,8,9] ]

# Rotate matrix 90 degrees...
for i in map(None,*original[::-1]):
    print str(i) + '\n'

这导致双方旋转90度(即。123(上面)现在是741(左边)。

这个Python解决方案是可行的,因为它使用了带负步的切片来反转行顺序(将7移到最上面)

original = [ [7,8,9],
             [4,5,6],
             [1,2,3] ]

然后,它使用map(以及隐含的标识函数,这是map以None作为第一个参数的结果)和*按顺序解包所有元素,重新组合列(即。第一个元素一起放在一个元组中,第二个元素一起放在一个元组中,以此类推)。你有效地得到得到返回如下重组:

original = [[7,8,9],
             [4,5,6],
             [1,2,3]]

下面是一个c#静态泛型方法,它可以为您完成这项工作。变量的名称很好,所以您可以很容易地理解算法的思想。

private static T[,] Rotate180 <T> (T[,] matrix)
{
    var height = matrix.GetLength (0);
    var width = matrix.GetLength (1);
    var answer = new T[height, width];

    for (int y = 0; y < height / 2; y++)
    {
        int topY = y;
        int bottomY = height - 1 - y;
        for (int topX = 0; topX < width; topX++)
        {
            var bottomX = width - topX - 1;
            answer[topY, topX] = matrix[bottomY, bottomX];
            answer[bottomY, bottomX] = matrix[topY, topX];
        }
    }

    if (height % 2 == 0)
        return answer;

    var centerY = height / 2;
    for (int leftX = 0; leftX < Mathf.CeilToInt(width / 2f); leftX++)
    {
        var rightX = width - 1 - leftX;
        answer[centerY, leftX] = matrix[centerY, rightX];
        answer[centerY, rightX] = matrix[centerY, leftX];
    }

    return answer;
}

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)

为新手程序员,在纯c++。(宝蓝的东西)

#include<iostream.h>
#include<conio.h>

int main()
{
    clrscr();

    int arr[10][10];        // 2d array that holds input elements 
    int result[10][10];     //holds result

    int m,n;                //rows and columns of arr[][]
    int x,y;                //rows and columns of result[][]

    int i,j;                //loop variables
    int t;                  //temporary , holds data while conversion

    cout<<"Enter no. of rows and columns of array: ";
    cin>>m>>n;
    cout<<"\nEnter elements of array: \n\n";
    for(i = 0; i < m; i++)
    {
        for(j = 0; j<n ; j++)
        {
          cin>>arr[i][j];         // input array elements from user
        }
    }


   //rotating matrix by +90 degrees

    x = n ;                      //for non-square matrix
    y = m ;     

    for(i = 0; i < x; i++)
    {  t = m-1;                     // to create required array bounds
       for(j = 0; j < y; j++)
       {
          result[i][j] = arr[t][i];
          t--;
       }
   }

   //print result

   cout<<"\nRotated matrix is: \n\n";
   for(i = 0; i < x; i++)
   {
       for(j = 0; j < y; j++)
       {
             cout<<result[i][j]<<" ";
       }
       cout<<"\n";
   }

   getch();
   return 0;
}