linear-list/array/rotate-image

Rotate Image

描述

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

Rotate the image by 90 degrees (clockwise).

Follow up: Could you do this in-place?

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。

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

// Rotate Image
// 思路 1，时间复杂度O(n^2)，空间复杂度O(1)
public class Solution {
    public void rotate(final int[][] matrix) {
        final int n = matrix.length;

        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, 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, n - 1 - i, j);
    }
    private static void swap(final int[][] matrix,
            int i, int j, int p, int q) {
        int tmp = matrix[i][j];
        matrix[i][j] = matrix[p][q];
        matrix[p][q] = tmp;
    }
};

// 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

// Rotate Image
// 思路 2，时间复杂度O(n^2)，空间复杂度O(1)
public class Solution {
    public void rotate(final int[][] matrix) {
        final int n = matrix.length;

        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, 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, j, i);
    }
    private static void swap(final int[][] matrix,
            int i, int j, int p, int q) {
        int tmp = matrix[i][j];
        matrix[i][j] = matrix[p][q];
        matrix[p][q] = tmp;
    }
};

// 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]);
    }
};