Question
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]
]
Solution 1
There are two major solutions to the problem. The most obvious idea is to rotate the image from outside to inside.
Code
Here is a sample solution.
class Solution {
public void rotate(int[][] matrix) {
int s = 0;
int e = matrix.length-1;
while(s<e) {
int len = e-s;
for(int i = 0; i<len; i++) {
int tmp = matrix[s][s+i];
matrix[s][s+i] = matrix[e-i][s];
matrix[e-i][s] = matrix[e][e-i];
matrix[e][e-i] = matrix[s+i][e];
matrix[s+i][e] = tmp;
}
s++;
e--;
}
}
}
Solution 2
Another solution is to firstly transpose the image and then flip it symmetrically. For instance:
First reverse up to down, then swap the symmetry
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
Code
Here is a sample solution.
public class Solution {
public void rotate(int[][] matrix) {
for(int i = 0; i<matrix.length; i++){
for(int j = i; j<matrix[0].length; j++){
int temp = matrix[i][j];
matrix[i][j] = matrix[j][i];
matrix[j][i] = temp;
}
}
for(int i =0 ; i<matrix.length; i++){
for(int j = 0; j<matrix.length/2; j++){
int temp = matrix[i][j];
matrix[i][j] = matrix[i][matrix.length-1-j];
matrix[i][matrix.length-1-j] = temp;
}
}
}
}
Performance
Both solutions are O(n x n) time complexity and use constant extra space.