C++实现LeetCode(48.旋转图像)

2024-12-09 19:45:40

[LeetCode] 48. Rotate Image 旋转图像

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.

Example 1:

Given input matrix =
[
[1,2,3],
[4,5,6],
[7,8,9]
],

rotate the input matrix in-place such that it becomes:
[
[7,4,1],
[8,5,2],
[9,6,3]
]

Example 2:

Given input matrix =
[
[ 5, 1, 9,11],
[ 2, 4, 8,10],
[13, 3, 6, 7],
[15,14,12,16]
],

rotate the input matrix in-place such that it becomes:
[
[15,13, 2, 5],
[14, 3, 4, 1],
[12, 6, 8, 9],
[16, 7,10,11]
]

在计算机图像处理里，旋转图片是很常见的，由于图片的本质是二维数组，所以也就变成了对数组的操作处理，翻转的本质就是某个位置上数移动到另一个位置上，比如用一个简单的例子来分析：

1 2 3　　　　　 7 4 1　

4 5 6　　-->　　 8 5 2　　

7 8 9 　　　　　9 6 3

对于90度的翻转有很多方法，一步或多步都可以解，先来看一种直接的方法，这种方法是按顺时针的顺序去覆盖前面的数字，从四个顶角开始，然后往中间去遍历，每次覆盖的坐标都是同理，如下：

(i, j) <- (n-1-j, i) <- (n-1-i, n-1-j) <- (j, n-1-i)

这其实是个循环的过程，第一个位置又覆盖了第四个位置，这里i的取值范围是 [0, n/2)，j的取值范围是 [i, n-1-i)，至于为什么i和j是这个取值范围，为啥i不用遍历 [n/2, n)，若仔细观察这些位置之间的联系，不难发现，实际上j列的范围 [i, n-1-i) 顺时针翻转 90 度，正好就是i行的 [n/2, n) 的位置，这个方法每次循环换四个数字，如下所示：

1 2 3               7 2 1            7 4 1

4 5 6      -->      4 5 6　　 --> 　 8 5 2　　

7 8 9               9 8 3　　　　   　 9 6 3

解法一：

class Solution {
public:
    void rotate(vector<vector<int>>& matrix) {
        int n = matrix.size();
        for (int i = 0; i < n / 2; ++i) {
            for (int j = i; j < n - 1 - i; ++j) {
                int tmp = matrix[i][j];
                matrix[i][j] = matrix[n - 1 - j][i];
                matrix[n - 1 - j][i] = matrix[n - 1 - i][n - 1 - j];
                matrix[n - 1 - i][n - 1 - j] = matrix[j][n - 1 - i];
                matrix[j][n - 1 - i] = tmp;
            }
        }
    }
};

还有一种解法，首先以从对角线为轴翻转，然后再以x轴中线上下翻转即可得到结果，如下图所示(其中蓝色数字表示翻转轴)：

1 2 3　　　　　 9 6 3　　　　　     7 4 1

4 5 6　　-->　　 8 5 2　　 -->   　 8 5 2　　

7 8 9 　　　　　7 4 1　　　　　     9 6 3

解法二：

class Solution {
public:
    void rotate(vector<vector<int>>& matrix) {
        int n = matrix.size();
        for (int i = 0; i < n - 1; ++i) {
            for (int j = 0; j < n - i; ++j) {
                swap(matrix[i][j], matrix[n - 1- j][n - 1 - i]);
            }
        }
        reverse(matrix.begin(), matrix.end());
    }
};

最后再来看一种方法，这种方法首先对原数组取其转置矩阵，然后把每行的数字翻转可得到结果，如下所示（其中蓝色数字表示翻转轴，Github 上可能无法显示颜色，请参见博客园上的帖子）：

1 2 3　　　　　 1 4 7　　　　　 7 4 1

4 5 6　　-->　　 2 5 8　　 --> 　 8 5 2　　

7 8 9 　　　　　3 6 9　　　　 9 6 3

解法三：

class Solution {
public:
    void rotate(vector<vector<int>>& matrix) {
        int n = matrix.size();
        for (int i = 0; i < n; ++i) {
            for (int j = i + 1; j < n; ++j) {
                swap(matrix[i][j], matrix[j][i]);
            }
            reverse(matrix[i].begin(), matrix[i].end());
        }
    }
};

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