48. Rotate Image
problem description
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]
]
algorithm thought
最简单的方法肯定是定义一个矩阵,然后将这个矩阵转置之后赋值给这个矩阵。但是这样需要O(n²)空间题目也明显要求了我们空间复杂度维持在O(1)内。我们必须在本来的数组上操作。如果只能这么操作的话,肯定就是使用swap函数,交换值。我们可以对称的改变矩阵,对于旋转90度,我们只需要首先按照竖直中线对称,再按照对角线对称就能得到结果。
code
class Solution {
public:
void rotate(vector<vector<int>>& matrix) {
int n=matrix.size();
for(int i=0;i<n/2;i++)
for(int j=0;j<n;++j){
swap(matrix[i][j],matrix[n-1-i][j]);
}
for(int i=0;i<n;i++)
for(int j=i+1;j<n;++j){
swap(matrix[i][j],matrix[j][i]);
}
}
};
algorithm analysis
对二维矩阵操作,双重循环。最后时间复杂度是O(n²)
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