Rotate Image

阎晋
2023-12-01

1,题目要求

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]
]

您将获得表示图像的n x n 2D矩阵。

将图像旋转90度(顺时针)。

2,题目思路

对于这道题,要求的是将一个二维矩阵顺时针旋转90度。

如果我们可以使用额外空间,实际并不难,直接按顺序进行映射即可。
但是题目不许使用额外内存,于是就需要在原二维矩阵上进行操作。

这里,我们使用LeetCode中Discussion的经典回答:

  • 顺时针旋转
/*
 * clockwise rotate
 * first reverse up to down, then swap the symmetry 
 * 1 2 3     7 8 9     7 4 1
 * 4 5 6  => 4 5 6  => 8 5 2
 * 7 8 9     1 2 3     9 6 3
*/
void rotate(vector<vector<int> > &matrix) {
    reverse(matrix.begin(), matrix.end());
    for (int i = 0; i < matrix.size(); ++i) {
        for (int j = i + 1; j < matrix[i].size(); ++j)
            swap(matrix[i][j], matrix[j][i]);
    }
}
  • 逆时针旋转
/*
 * anticlockwise rotate
 * first reverse left to right, then swap the symmetry
 * 1 2 3     3 2 1     3 6 9
 * 4 5 6  => 6 5 4  => 2 5 8
 * 7 8 9     9 8 7     1 4 7
*/
void anti_rotate(vector<vector<int> > &matrix) {
    for (auto vi : matrix) reverse(vi.begin(), vi.end());
    for (int i = 0; i < matrix.size(); ++i) {
        for (int j = i + 1; j < matrix[i].size(); ++j)
            swap(matrix[i][j], matrix[j][i]);
    }
}

3,代码实现

static auto speedup = [](){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    return nullptr;
}();

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

相关阅读

相关文章

相关问答