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?
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
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++;
}
}
}
其他回答
#!/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]]
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;
}
很好的答案,但对于那些正在寻找DRY JavaScript代码的人- +90度和-90度:
// Input: 1 2 3 // 4 5 6 // 7 8 9 // Transpose: // 1 4 7 // 2 5 8 // 3 6 9 // Output: // +90 Degree: // 7 4 1 // 8 5 2 // 9 6 3 // -90 Degree: // 3 6 9 // 2 5 8 // 1 4 7 // Rotate +90 function rotate90(matrix) { matrix = transpose(matrix); matrix.map(function(array) { array.reverse(); }); return matrix; } // Rotate -90 function counterRotate90(matrix) { var result = createEmptyMatrix(matrix.length); matrix = transpose(matrix); var counter = 0; for (var i = matrix.length - 1; i >= 0; i--) { result[counter] = matrix[i]; counter++; } return result; } // Create empty matrix function createEmptyMatrix(len) { var result = new Array(); for (var i = 0; i < len; i++) { result.push([]); } return result; } // Transpose the matrix function transpose(matrix) { // make empty array var len = matrix.length; var result = createEmptyMatrix(len); for (var i = 0; i < matrix.length; i++) { for (var j = 0; j < matrix[i].length; j++) { var temp = matrix[i][j]; result[j][i] = temp; } } return result; } // Test Cases var array1 = [ [1, 2], [3, 4] ]; var array2 = [ [1, 2, 3], [4, 5, 6], [7, 8, 9] ]; var array3 = [ [1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16] ]; // +90 degress Rotation Tests var test1 = rotate90(array1); var test2 = rotate90(array2); var test3 = rotate90(array3); console.log(test1); console.log(test2); console.log(test3); // -90 degress Rotation Tests var test1 = counterRotate90(array1); var test2 = counterRotate90(array2); var test3 = counterRotate90(array3); console.log(test1); console.log(test2); console.log(test3);
这是c#的
int[,] array = new int[4,4] {
{ 1,2,3,4 },
{ 5,6,7,8 },
{ 9,0,1,2 },
{ 3,4,5,6 }
};
int[,] rotated = RotateMatrix(array, 4);
static int[,] RotateMatrix(int[,] matrix, int n) {
int[,] ret = new int[n, n];
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
ret[i, j] = matrix[n - j - 1, i];
}
}
return ret;
}
下面是我的Ruby版本(注意,值显示的不一样,但它仍然按照描述旋转)。
def rotate(matrix)
result = []
4.times { |x|
result[x] = []
4.times { |y|
result[x][y] = matrix[y][3 - x]
}
}
result
end
matrix = []
matrix[0] = [1,2,3,4]
matrix[1] = [5,6,7,8]
matrix[2] = [9,0,1,2]
matrix[3] = [3,4,5,6]
def print_matrix(matrix)
4.times { |y|
4.times { |x|
print "#{matrix[x][y]} "
}
puts ""
}
end
print_matrix(matrix)
puts ""
print_matrix(rotate(matrix))
输出:
1 5 9 3
2 6 0 4
3 7 1 5
4 8 2 6
4 3 2 1
8 7 6 5
2 1 0 9
6 5 4 3