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2014
11-19

LeetCode-Rotate Image[数组]

Rotate Image

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?

标签: Array
分析

首先想到,纯模拟,从外到内一圈一圈的转,但这个方法太慢。

如下图,首先沿着副对角线翻转一次,然后沿着水平中线翻转一次。

\begin{center}
\includegraphics[width=200pt]{rotate-image.png}\\
\figcaption{Rotate Image}\label{fig:rotate-image}
\end{center}

或者,首先沿着水平中线翻转一次,然后沿着主对角线翻转一次。

代码1

// LeetCode, Rotate Image
// 思路 1,时间复杂度O(n^2),空间复杂度O(1)
class Solution {
public:
    void rotate(vector<vector<int>>& matrix) {
        const int n = matrix.size();

        for (int i = 0; i < n; ++i)  // 沿着副对角线反转
            for (int j = 0; j < n - i; ++j)
                swap(matrix[i][j], matrix[n - 1 - j][n - 1 - i]);

        for (int i = 0; i < n / 2; ++i) // 沿着水平中线反转
            for (int j = 0; j < n; ++j)
                swap(matrix[i][j], matrix[n - 1 - i][j]);
    }
};

代码2

// LeetCode, Rotate Image
// 思路 2,时间复杂度O(n^2),空间复杂度O(1)
class Solution {
public:
    void rotate(vector<vector<int>>& matrix) {
        const int n = matrix.size();

        for (int i = 0; i < n / 2; ++i) // 沿着水平中线反转
            for (int j = 0; j < n; ++j)
                swap(matrix[i][j], matrix[n - 1 - i][j]);

        for (int i = 0; i < n; ++i)  // 沿着主对角线反转
            for (int j = i + 1; j < n; ++j)
                swap(matrix[i][j], matrix[j][i]);
    }
};