Algorithm
Problem Name: 48. Rotate Image
You are given an n x n 2D matrix representing an image, rotate the image by 90 degrees (clockwise).
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:
Input: matrix = [[1,2,3],[4,5,6],[7,8,9]] Output: [[7,4,1],[8,5,2],[9,6,3]]
Example 2:
Input: matrix = [[5,1,9,11],[2,4,8,10],[13,3,6,7],[15,14,12,16]] Output: [[15,13,2,5],[14,3,4,1],[12,6,8,9],[16,7,10,11]]
Constraints:
n == matrix.length == matrix[i].length1 <= n <= 20-1000 <= matrix[i][j] <= 1000
Code Examples
#1 Code Example with C Programming
Code -
C Programming
#define RIGHT(I) (matrix[I][y])
#define BOTTOM(I) (matrix[y][y + x - I])
#define LEFT(I) (matrix[y + x - I][x])
#define TOP(I) (matrix[x][I])
void spin(int **matrix, int x, int y) {
int i;
int a, b, c;
if (x >= y) return;
for (i = x; i < y; i ++) {
a = RIGHT(i);
RIGHT(i) = TOP(i);
b = BOTTOM(i);
BOTTOM(i) = a;
c = LEFT(i);
LEFT(i) = b;
TOP(i) = c;
}
spin(matrix, x + 1, y - 1);
}
void rotate(int** matrix, int matrixRowSize, int matrixColSize) {
spin(matrix, 0, matrixRowSize - 1);
}
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#2 Code Example with C++ Programming
Code -
C++ Programming
class Solution {
public:
void rotate(vector<vector<int>>& matrix) {
helper(matrix, 0, 0, matrix.size());
}
void helper(vector < vector<int>>& matrix, int row, int col, int size) {
if (size == 0 || size == 1) return;
int step = 0;
while(step < size - 1){
swap(matrix[row][col + step], matrix[row + step][col + size - 1]);
swap(matrix[row][col + step], matrix[row + size - 1 - step][col]);
swap(matrix[row + size - 1][col + size - 1 - step], matrix[row + size - 1 - step][col]);
step++;
}
helper(matrix, row + 1, col + 1, size - 2);
}
};
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Output
#3 Code Example with Java Programming
Code -
Java Programming
class Solution {
public void rotate(int[][] matrix) {
int n = matrix.length;
for (int i = 0; i < (n + 1) / 2; i++) {
for (int j = 0; j < n / 2; j++) {
int temp = matrix[n - 1 - j][i];
matrix[n - 1 - j][i] = matrix[n - 1 - i][n - j - 1];
matrix[n - 1 - i][n - j - 1] = matrix[j][n - 1 - i];
matrix[j][n - 1 - i] = matrix[i][j];
matrix[i][j] = temp;
}
}
}
}
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#4 Code Example with Javascript Programming
Code -
Javascript Programming
const rotate = function(matrix) {
let s = 0,
e = matrix.length - 1
while (s < e) {
let temp = matrix[s]
matrix[s] = matrix[e]
matrix[e] = temp
s++
e--
}
for (let i = 0; i < matrix.length; i++) {
for (let j = i + 1; j < matrix[i].length; j++) {
let temp = matrix[i][j]
matrix[i][j] = matrix[j][i]
matrix[j][i] = temp
}
}
}
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#5 Code Example with Python Programming
Code -
Python Programming
class Solution:
def rotate(self, matrix):
matrix[:] = [[row[i] for row in matrix[::-1]] for i in range(len(matrix))]
Input
Output