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] ]
一个n x n的二维矩阵表示一个图像，将图像顺时针旋转90度。要求in-place，所以就不能用额外的空间了。
解法1: 先以对角线为轴翻转得到其转置矩阵，再以中间竖轴翻转。
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2: 先以反对角线翻转，在以中间水平轴翻转。
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
Java：
public class Solution {
public void rotate(int[][] matrix) {
int n = matrix.length;
// along the left top to right bottom diagonal line, swap symmetrical pair
for(int i=0; i<n; i++) { // for each row
for(int j=i+1; j<n; j++) { // for each number
// swap the pair
int temp = matrix[i][j];
matrix[i][j] = matrix[j][i];
matrix[j][i] = temp;
}
}
// flip each row horizontally
for(int i=0; i<n; i++) {
for(int j=0; j<n/2; j++) {
int temp = matrix[i][j];
matrix[i][j] = matrix[i][n-1-j];
matrix[i][n-1-j] = temp;
}
}
}
}
Python: T: O(n^2), S: O(1)
class Solution:
# @param matrix, a list of lists of integers
# @return a list of lists of integers
def rotate(self, matrix):
n = len(matrix)
# anti-diagonal mirror
for i in xrange(n):
for j in xrange(n - i):
matrix[i][j], matrix[n-1-j][n-1-i] = matrix[n-1-j][n-1-i], matrix[i][j]
# horizontal mirror
for i in xrange(n / 2):
for j in xrange(n):
matrix[i][j], matrix[n-1-i][j] = matrix[n-1-i][j], matrix[i][j]
return matrix
C++:
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());
}
}
};