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[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());
        }
    }
};

 

Github 同步地址:

https://github.com/grandyang/leetcode/issues/48

 

参考资料:

https://leetcode.com/problems/rotate-image/

https://leetcode.com/problems/rotate-image/discuss/18895/Clear-Java-solution

https://leetcode.com/problems/rotate-image/discuss/18872/A-common-method-to-rotate-the-image

 

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posted @ 2015-04-03 11:33  Grandyang  阅读(...)  评论(... 编辑 收藏