You are given an n x n 2D matrix representing an image.

Rotate the image by 90 degrees (clockwise).

Note:

You have to rotate the image in-place, which means you have to modify the input 2D matrix directly. DO NOT allocate another 2D matrix and do the rotation.

Example 1:

Given input matrix = 
[
  [1,2,3],
  [4,5,6],
  [7,8,9]
],

rotate the input matrix in-place such that it becomes:
[
  [7,4,1],
  [8,5,2],
  [9,6,3]
]

Example 2:

Given input matrix =
[
  [ 5, 1, 9,11],
  [ 2, 4, 8,10],
  [13, 3, 6, 7],
  [15,14,12,16]
], 

rotate the input matrix in-place such that it becomes:
[
  [15,13, 2, 5],
  [14, 3, 4, 1],
  [12, 6, 8, 9],
  [16, 7,10,11]
]

一个n x n的二维矩阵表示一个图像，将图像顺时针旋转90度。要求in-place，所以就不能用额外的空间了。

解法1: 先以对角线为轴翻转得到其转置矩阵，再以中间竖轴翻转。

1 2 3　　　　1 4 7　　　　 7 4 1

4 5 6　-->　 2 5 8　 --> 8 5 2　　

7 8 9 　　　 3 6 9　　　9 6 3

解法2: 先以反对角线翻转，在以中间水平轴翻转。

1 2 3　　　　9 6 3　　　　7 4 1

4 5 6　-->　 8 5 2　 --> 8 5 2　　

7 8 9 　　　 7 4 1　　　　 9 6 3

Java：

public class Solution {
    public void rotate(int[][] matrix) {
        int n = matrix.length;
        
        // along the left top to right bottom diagonal line, swap symmetrical pair
        for(int i=0; i<n; i++) {  // for each row
            for(int j=i+1; j<n; j++) {  // for each number
                // swap the pair
                int temp = matrix[i][j];
                matrix[i][j] = matrix[j][i];
                matrix[j][i] = temp;
            }
        }
        
        // flip each row horizontally 
        for(int i=0; i<n; i++) {
            for(int j=0; j<n/2; j++) {
                int temp = matrix[i][j];
                matrix[i][j] = matrix[i][n-1-j];
                matrix[i][n-1-j] = temp;
                
            }
        }
    }
}

Python: T: O(n^2), S: O(1)

class Solution:
    # @param matrix, a list of lists of integers
    # @return a list of lists of integers
    def rotate(self, matrix):
        n = len(matrix)
        
        # anti-diagonal mirror
        for i in xrange(n):
            for j in xrange(n - i):
                matrix[i][j], matrix[n-1-j][n-1-i] = matrix[n-1-j][n-1-i], matrix[i][j]
        
        # horizontal mirror
        for i in xrange(n / 2):
            for j in xrange(n):
                matrix[i][j], matrix[n-1-i][j] = matrix[n-1-i][j], matrix[i][j]
                
        return matrix

C++:

class Solution {
public:
    void rotate(vector<vector<int> > &matrix) {
        int n = matrix.size();
        for (int i = 0; i < n; ++i) {
            for (int j = i + 1; j < n; ++j) {
                swap(matrix[i][j], matrix[j][i]);
            }
            reverse(matrix[i].begin(), matrix[i].end());
        }
    }
};