【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:

【每日一题】48. Rotate Image_第1张图片
img
Input: matrix = [[1,2,3],[4,5,6],[7,8,9]]
Output: [[7,4,1],[8,5,2],[9,6,3]]

Example 2:

【每日一题】48. Rotate Image_第2张图片
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>& matrix) {
        if (matrix.size() <= 0) return;
        updown(matrix);
        diagonal(matrix);
        
        return;
    }
    void updown(vector>& 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>& 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]);
            }
        }
    }
};

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