Problem description:

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.

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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]
]

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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:

  • clockwise rotate:
    first reverse up to down, then swap the symmetry 
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    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
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void rotate(vector<vector<int> > &matrix) {
    reverse(matrix.begin(), matrix.end());
    for (int i = 0; i < matrix.size(); ++i) {
        for (int j = i + 1; j < matrix[i].size(); ++j)
            swap(matrix[i][j], matrix[j][i]);
    }
}

Since we only need to do three swaps in 3*3 example, the following code is more readable

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

  • anticlockwise rotate:
    first reverse left to right, then swap the symmetry
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    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
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void anti_rotate(vector<vector<int> > &matrix) {
    for (auto vi : matrix) reverse(vi.begin(), vi.end());
    for (int i = 0; i < matrix.size(); ++i) {
        for (int j = i + 1; j < matrix[i].size(); ++j)
            swap(matrix[i][j], matrix[j][i]);
    }
}

time complexity: $O(n^2)$
space complexity: $O(1)$

Another solution:

The idea is to define the a range to do the swapping every loop.

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class Solution {
public:
    void rotate(vector<vector<int>>& matrix) {

        int top= 0, left= 0, bot= matrix.size()-1, right= matrix[0].size()-1;
        int n= matrix.size();
        while(n> 1){
            for(int i= 0; i< n-1; i++){
                int tmp= matrix[top][left+i];
                matrix[top][left+i]= matrix[bot-i][left];
                matrix[bot-i][left]= matrix[bot][right-i];
                matrix[bot][right-i]= matrix[top+i][right];
                matrix[top+i][right]= tmp;
            }
            top++; left++; right--; bot--;
            n-=2;
        }
        
    }
};

reference:
https://goo.gl/qnuhzL