【LeetCode从零单刷】Rotate Image

题目:

You are given an n x n 2D matrix representing an image.

Rotate the image by 90 degrees (clockwise).

Follow up:
Could you do this in-place?

解答:

一开始想的很复杂,希望找到一个通式应对所有角度的旋转。所有的细节写在《【图像处理】基于OpenCV底层实现的图片旋转》。

其实没有必要,因为90度是非常特殊的一种旋转,可以直观上赋值:

class Solution {
public:
    void rotate(vector<vector<int>>& matrix) {
        int m = matrix.size();
        int n = matrix[0].size();
        vector<int> tmp(m, 0);
        vector<vector<int>> ans(n, tmp);
        for(int i=0; i<m; i++)
            for(int j=0; j<n; j++)
            {
                ans[j][m-1 - i] = matrix[i][j];
            }
        matrix = ans;
        return;
    }
};

但仔细看题目要求:in-place(原地)。想了半天没想出来原地旋转的方法。直到看到巧妙做法:传送门。精粹就在下图中:

【LeetCode从零单刷】Rotate Image_第1张图片

class Solution {
public:
    void rotate(vector<vector<int> > &matrix) {
        int i,j,temp;
        int n=matrix.size();
        // 沿着副对角线反转
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n - i; ++j) {
                temp = matrix[i][j];
                matrix[i][j] = matrix[n - 1 - j][n - 1 - i];
                matrix[n - 1 - j][n - 1 - i] = temp;
            }
        }
        // 沿着水平中线反转
        for (int i = 0; i < n / 2; ++i){
            for (int j = 0; j < n; ++j) {
                temp = matrix[i][j];
                matrix[i][j] = matrix[n - 1 - i][j];
                matrix[n - 1 - i][j] = temp;
            }
        }
    }
};

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