【每日一题】48. Rotate Image

问题描述

You are given an n x n 2D matrix representing an image, rotate the image by 90 degrees (clockwise).

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:

img

Input: matrix = [[1,2,3],[4,5,6],[7,8,9]]
Output: [[7,4,1],[8,5,2],[9,6,3]]

Example 2:

img

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

Example 3:

Input: matrix = [[1]]
Output: [[1]]

Example 4:

Input: matrix = [[1,2],[3,4]]
Output: [[3,1],[4,2]]

Constraints:

• matrix.length == n
• matrix[i].length == n
• 1 <= n <= 20
• -1000 <= matrix[i][j] <= 1000

给定一个 n × n 的二维矩阵表示一个图像。

将图像顺时针旋转 90 度。

题解

给定一个方阵，将方阵顺时针旋转90度；而且要求必须原地选择，直接对矩阵内容进行修改，不能使用别的矩阵进行辅助。

通过观察，可以发现，顺时针矩阵旋转可以通过两步完成：

1. 上下翻转
2. 主对角线翻转。

image-20201028092105741

完整代码：

class Solution {
public:
    void rotate(vector<vector<int>>& matrix) {
        if (matrix.size() <= 0) return;
        updown(matrix);
        diagonal(matrix);
        
        return;
    }
    void updown(vector<vector<int>>& matrix){
        int size = matrix.size(), t = matrix.size() / 2;
        for (int i=0; i< t; ++i){
            for (int j=0; j< size; ++j){
                swap(matrix[i][j], matrix[size-1-i][j]);
            }
        }
    }
    void diagonal(vector<vector<int>>& matrix){
        int size = matrix.size();
        for (int i=0; i< size; ++i){
            for (int j=i; j< size; ++j){
                swap(matrix[i][j], matrix[j][i]);
            }
        }
    }
};