LeetCode //C - 48. Rotate Image-Toy模板网

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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:

LeetCode //C - 48. Rotate Image,LeetCode,leetcode,c语言,算法

Input: matrix = [[1,2,3],[4,5,6],[7,8,9]]
Output: [[7,4,1],[8,5,2],[9,6,3]]

Example 2:

LeetCode //C - 48. Rotate Image,LeetCode,leetcode,c语言,算法

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

From: LeetCode
Link: 48. Rotate Image

Solution:

Ideas：

To rotate a matrix by 90 degrees clockwise in place, we can follow a two-step process:

Transpose the Matrix: The transpose of a matrix is obtained by swapping rows and columns. This means
matrix[i][j] will become matrix[j][i].
Reverse the Rows: Reverse each row of the transposed matrix.

Following these steps will give us the rotated matrix.文章来源地址https://www.toymoban.com/news/detail-630395.html

Code:

void rotate(int** matrix, int matrixSize, int* matrixColSize) {
    // Transpose the matrix
    for (int i = 0; i < matrixSize; i++) {
        for (int j = i; j < matrixSize; j++) {
            int temp = matrix[i][j];
            matrix[i][j] = matrix[j][i];
            matrix[j][i] = temp;
        }
    }

    // Reverse each row
    for (int i = 0; i < matrixSize; i++) {
        for (int j = 0, k = matrixSize - 1; j < k; j++, k--) {
            int temp = matrix[i][j];
            matrix[i][j] = matrix[i][k];
            matrix[i][k] = temp;
        }
    }
}