1.新建一個數組，將原數組的數據按規律複製到新數組，這種方法做不到in-place，佔用了額外一個數組的空間
newx = y;
newy = n-1-x;2.我們可以按ring by ring的順序進行操作
交換在每個ring上的4個點之間進行
public class Solution {
public void rotate(int[][] matrix) {
int n = matrix.length;
for(int x = 0; x <= (n - 1) >> 1; x++) {
for(int y = x; y <= n - 2 - x; y++) {
int newx1 = y;
int newy1 = n - 1 - x;
int newx2 = newy1;
int newy2 = n - 1 - newx1;
int newx3 = newy2;
int newy3 = n - 1 - newx2;
int temp = matrix[newx1][newy1];
matrix[newx1][newy1] = matrix[x][y];
matrix[x][y] = matrix[newx3][newy3];
matrix[newx3][newy3] = matrix[newx2][newy2];
matrix[newx2][newy2] = temp;
}
}
}
}
3.將上下的行進行反轉，然後按主對角線進行對稱交換；這這種方法也很容易做到逆時針旋轉。
/*
* clockwise rotate
* first reverse up to down, then swap the symmetry
* 1 2 3 7 8 9 7 4 1
* 4 5 6 => 4 5 6 => 8 5 2
* 7 8 9 1 2 3 9 6 3
*/
/*
* anticlockwise rotate
* first reverse left to right, then swap the symmetry
* 1 2 3 3 2 1 3 6 9
* 4 5 6 => 6 5 4 => 2 5 8
* 7 8 9 9 8 7 1 4 7
*/