Rotate Image — LeetCode #48

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].length
• 1 <= n <= 20
• -1000 <= matrix[i][j] <= 1000

Solutions:

Python

class Solution:
    def rotate(self, matrix: List[List[int]]) -> None:
        """
        Do not return anything, modify matrix in-place instead.
        """
        # Transpose the matrix
        for i in range(len(matrix)):
            for j in range(i, len(matrix[0])):
                matrix[i][j], matrix[j][i] = matrix[j][i], matrix[i][j]
        
        # Reverse each row in the matrix
        for row in matrix:
            row.reverse()

C#

public class Solution {
    public void Rotate(int[][] matrix) {
        // Transpose the 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;
            }
        }

        // Reverse each row in the matrix
        for (int i = 0; i < matrix.Length; i++) {
            Array.Reverse(matrix[i]);
        }
    }
}

Java

class Solution {
    public void rotate(int[][] matrix) {
        // Transpose the 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;
            }
        }

        // Reverse each row in the matrix
        for (int i = 0; i < matrix.length; i++) {
            for (int j = 0; j < matrix[0].length / 2; j++) {
                int temp = matrix[i][j];
                matrix[i][j] = matrix[i][matrix[0].length - 1 - j];
                matrix[i][matrix[0].length - 1 - j] = temp;
            }
        }
    }
}

Javascript

/**
 * @param {number[][]} matrix
 * @return {void} Do not return anything, modify matrix in-place instead.
 */
var rotate = function(matrix) {
    // Transpose the matrix
    for (let i = 0; i < matrix.length; i++) {
        for (let j = i; j < matrix[0].length; j++) {
            [matrix[i][j], matrix[j][i]] = [matrix[j][i], matrix[i][j]];
        }
    }

    // Reverse each row in the matrix
    for (let row of matrix) {
        row.reverse();
    }
};

Typescript

/**
 Do not return anything, modify matrix in-place instead.
 */
function rotate(matrix: number[][]): void {
    // Transpose the matrix
    for (let i = 0; i < matrix.length; i++) {
        for (let j = i; j < matrix[0].length; j++) {
            [matrix[i][j], matrix[j][i]] = [matrix[j][i], matrix[i][j]];
        }
    }

    // Reverse each row in the matrix
    for (let row of matrix) {
        row.reverse();
    }
};

PHP

class Solution {

    /**
     * @param Integer[][] $matrix
     * @return NULL
     */
    function rotate(&$matrix) {
        // Transpose the matrix
        for ($i = 0; $i < count($matrix); $i++) {
            for ($j = $i; $j < count($matrix[0]); $j++) {
                $temp = $matrix[$i][$j];
                $matrix[$i][$j] = $matrix[$j][$i];
                $matrix[$j][$i] = $temp;
            }
        }

        // Reverse each row in the matrix
        for ($i = 0; $i < count($matrix); $i++) {
            for ($j = 0; $j < count($matrix[0]) / 2; $j++) {
                $temp = $matrix[$i][$j];
                $matrix[$i][$j] = $matrix[$i][count($matrix[0]) - 1 - $j];
                $matrix[$i][count($matrix[0]) - 1 - $j] = $temp;
            }
        }
    }
}

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