You are given an n x n 2D matrix representing an image.
Rotate the image by 90 degrees (clockwise).
Follow up:
Could you do this in-place? 来源:
分析
方法1:
关键在于找到旋转90°的关系
假设
a:(i,j),那么
b: (j,n-1-i)
c: (n-1-j,i)
d: (n-1-i,n-1-j)
依次交换就是了,循环的方式是:
从最外层一圈操作,然后往内圈重复步骤
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 | 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; } } } }; |
方法2
先将矩阵转置,然后对每一行逆序
1 2 3 4 5 6 7 8 9 10 11 12 | class Solution { public : void rotate(vector<vector< int >>& m) { int n = m.size(); for ( int i=0; i<n; i++) for ( int j=0; j<i; j++) swap(m[i][j], m[j][i]); for ( int i=0; i<n; i++) reverse(m[i].begin(), m[i].end()); } }; |
类似的,如果想要得到逆时针90度,那么就先对每一行求逆,然后再转置