矩阵旋转的算法

对于 n*n矩阵旋转，取矩阵元素时从下标0开始，则(x,y)旋转后的位置为(y,n-1-x)，在不考虑空间的情况下，可以遍历矩阵，根据旋转关系将元素填入新矩阵对应位置即可

伪代码为

for i = 0 to n -1

     for j = 0 to n - 1

         b[i][j] = a[j][n - 1 - i]

这种算法的空间复杂度为O(n*n)

另一种方法，空间复杂度为O(1)

假设n=4

最外层的旋转路线为:

(0,0)->(0,3)->(3,3)->(3,0)->(0,0)

(0,1)->(1,3)->(3,2)->(2,0)->(0,1)

(0,2)->(2,3)->(3,1)->(1,0)->(0,2)

次层旋转路线为:

(1,1)->(1,2)->(2,2)->(2, 1)->(1,1)

依此类推。

代码如下

#include <iostream>
#include <iomanip>

using namespace std;

void rotate(int** matrix, int matrixRowSize, int matrixColSize)
{
int n = matrixRowSize;
int m = n / 2;

for (int i = 0; i < m; i++) {
for (int j = i; j < n - 1 - i; j++) {
int startx = i, starty = j;
int x = startx, y = starty;
int newx = y, newy = n - 1 - x;
int tmp1 = *((int*)matrix + startx * matrixColSize + starty);
int tmp2;

while (newx != startx || newy != starty) {
tmp2 = *((int*)matrix + newx * matrixColSize + newy);
*((int*)matrix + newx * matrixColSize + newy) = tmp1;
x = newx, y = newy;
newx = y, newy = n - 1 - x;
tmp1 = tmp2;
}

*((int*)matrix + startx * matrixColSize + starty) = tmp1;
}
}

}

int main()
{
const int n = 2;
int matrix

= { { 1, 2}, { 3, 4} };

cout << "before rotate:" << endl;
for (int i = 0; i < n; i++)
{
for (int j = 0; j < n; j++)
{
cout << right << setw(3) << matrix[i][j];
}
cout << endl;
}

rotate((int**)matrix, n, n);

cout << "after rotate:" << endl;
for (int i = 0; i < n; i++)
{
for (int j = 0; j < n; j++)
{
cout << right << setw(3) << matrix[i][j];
}
cout << endl;
}
return 0;
}