[leetcode]48. Rotate Image

最新推荐文章于 2025-08-14 23:25:10 发布

你看见我的代码了么

最新推荐文章于 2025-08-14 23:25:10 发布

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分类专栏： LeetCode Medium 文章标签： leetcode

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本文详细解析了LeetCode上第48题“Rotate Image”的解决方案，介绍了一种通过先翻转再交换对称元素来实现矩阵顺时针旋转90度的高效算法，并提供了逆时针旋转的类似方法。

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[leetcode]48. Rotate Image

Analysis

好冷鸭—— [每天刷题并不难0.0]

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.
在这里插入图片描述
矩阵的旋转

Implement

class Solution {
public:
    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]);
        }
    }
};

下面是矩阵旋转的通用方法

clockwise

/*
 * clockwise rotate
 * first reverse up to down, then swap the symmetry 
 * 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
*/
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]);
    }
}

anticlockwise

/*
 * anticlockwise rotate
 * first reverse left to right, then swap the symmetry
 * 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
*/
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]);
    }
}