# 48. Rotate Image

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]
]

# Solution

Approach 1: For each range (e.g., 1, 2, 3, 6, 9, 8, 7, 4), rotate each circle (e.g., 1, 3, 9, 7).

Approach 2: Transpose the matrix and then reverse columns.

# Code (Python)

Approach 1:

Approach 2:

# Code (C++)

Approach 1:

// e.g., 1 2 3
//       4 5 6
//       7 8 9
// For each range (e.g., 1, 2, 3, 6, 9, 8, 7, 4), rotate each circle (e.g., 1, 3, 9, 7).
class Solution {
private:
    vector<pair<int,int>> steps = {{0,1}, {1,0}, {0,-1}, {-1,0}};
public:
    void rotate(vector<vector<int>>& matrix) {
        int n = matrix.size();
        for (int i = 0; i < n / 2; ++i)
        {
            int len = n - 2 * i;
            for (int j = i; j < i + len - 1; ++j)
            {
                int row = i;
                int col = j;
/*
                int stepIdx = 0;
                int tmp = matrix[row][col];
                do
                {
                    for (int k = 0; k < len - 1; ++k)
                    {
                        int nextRow = row + steps[stepIdx].first;
                        int nextCol = col + steps[stepIdx].second;
                        if (nextRow < i || nextRow >= i + len || nextCol < i || nextCol >= i + len)
                            stepIdx = (stepIdx + 1) % steps.size();
                        row += steps[stepIdx].first;
                        col += steps[stepIdx].second;
                    }
                    std::swap(tmp, matrix[row][col]);
                } while (row != i || col != j);
*/
                int tmp[4];
                for (int k = 0; k < 4; ++k)
                {
                    tmp[k] = matrix[row][col];
                    int tmpRow = row;
                    row = col;
                    col = n - tmpRow - 1;
                }
                for (int k = 0; k < 4; ++k)
                {
                    matrix[row][col] = tmp[(k + 3) % 4];
                    int tmpRow = row;
                    row = col;
                    col = n - tmpRow - 1;
                }
            }
        }
    }
};

Approach 2:

// e.g., 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
class Solution {
public:
    void rotate(vector<vector<int>>& matrix) {
        int n = matrix.size();
        for (int i = 0; i < n; ++i)
        {
            for (int j = 0; j < i; ++j)
            {
                std::swap(matrix[i][j], matrix[j][i]);
            }
        }
        for (int i = 0; i < n; ++i)
        {
            for (int j = 0; j < n / 2; ++j)
            {
                std::swap(matrix[i][j], matrix[i][n-j-1]);
            }
        }
    }
};