LeetCode 刷题记录 48. Rotate Image

题目：
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]
]

首先可以沿着对角线翻转即进行转置操作，然后沿着中间的列进行列翻转

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

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]);
            }
        }
        for(int i = 0; i < n; i++){
            for(int j = 0; j < n / 2; j++){
                swap(matrix[i][j],matrix[i][n-1-j]);
            }
        }
    }
};

java：

class Solution {
    public void rotate(int[][] matrix) {
        int n = matrix.length;
        for(int i = 0; i < n; i++){
            for(int j = i + 1; j < n; j++){
                int temp = matrix[i][j];
                matrix[i][j] = matrix[j][i];
                matrix[j][i] = temp;
            }
        }
        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：

class Solution(object):
    def rotate(self, matrix):
        """
        :type matrix: List[List[int]]
        :rtype: None Do not return anything, modify matrix in-place instead.
        """
        n = len(matrix)
        
        for i in xrange(n):
            for j in xrange(i + 1, n):
                matrix[i][j],matrix[j][i] = matrix[j][i],matrix[i][j]
                
        for i in xrange(n):
            for j in xrange(n/2):
                matrix[i][j],matrix[i][n - 1 - j] = matrix[i][n - 1- j],matrix[i][j]

另外也可以先根据副对角线翻转，然后上下翻转