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

对于i:= 0到X 对于j:= 0到X做 [j][i]:= graphic2[X-i][j]

X是图形所在数组的大小。

其他回答

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);

}

}

    public static void rotateMatrix(int[,] matrix)
    {
        //C#, to rotate an N*N matrix in place
        int n = matrix.GetLength(0);
        int layers =  n / 2;
        int temp, temp2;

        for (int i = 0; i < layers; i++) // for a 5 * 5 matrix, layers will be 2, since at layer three there would be only one element, (2,2), and we do not need to rotate it with itself 
        {
            int offset = 0;
            while (offset < n - 2 * i - 1)
            {
                // top right <- top left 
                temp = matrix[i + offset, n - i - 1]; //top right value when offset is zero
                matrix[i + offset, n - i - 1] = matrix[i, i + offset];   

                //bottom right <- top right 
                temp2 = matrix[n - i - 1, n - i - 1 - offset]; //bottom right value when offset is zero
                matrix[n - i - 1, n - i - 1 - offset] = temp;  

                //bottom left <- bottom right 
                temp = matrix[n - i - 1 - offset, i];
                matrix[n - i - 1 - offset, i] = temp2;  

                //top left <- bottom left 
                matrix[i, i + offset] = temp; 

                offset++;
            }
        }
    }

时间- O(N),空间- O(1)

public void rotate(int[][] matrix) {
    int n = matrix.length;
    for (int i = 0; i < n / 2; i++) {
        int last = n - 1 - i;
        for (int j = i; j < last; j++) {
            int top = matrix[i][j];
            matrix[i][j] = matrix[last - j][i];
            matrix[last - j][i] = matrix[last][last - j];
            matrix[last][last - j] = matrix[j][last];
            matrix[j][last] = top;
        }
    }
}

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)

下面是Java版本:

public static void rightRotate(int[][] matrix, int n) {
    for (int layer = 0; layer < n / 2; layer++) {
        int first = layer;
        int last = n - 1 - first;
        for (int i = first; i < last; i++) {
           int offset = i - first;
           int temp = matrix[first][i];
           matrix[first][i] = matrix[last-offset][first];
           matrix[last-offset][first] = matrix[last][last-offset];
           matrix[last][last-offset] = matrix[i][last];
           matrix[i][last] = temp;
        }
    }
}

该方法首先旋转最外层,然后按顺序移动到内层。