C language force buckles the rotating image of the 48th question.auxiliary array

Given a n × n two-dimensional matrix matrix representing an image.Please rotate the image 90 degrees clockwise.

You must be in in-place rotates the image, which means you need to modify the input 2D matrix directly.Please do not use another matrix to rotate the image.

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]]

Tip:

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

Method ideas:

The idea is exactly the same as the matrix transpose, and if you are interested, you can refer to another blog of minePostgraduate Review ReviewNotes on Data Structures (7) Arrays and Strings (Part 1) (including related content of arrays) - Programmer Sought

Compressed storage of matrices

We use the second example in the title

As an example, analyze where the numbers appear after rotating the image 90 degrees.

For the first row in the matrix, after rotation, it appears in the penultimate column position:

And, the xth element of the first row is exactly the xth element of the penultimate column after rotation.

For the second row in the matrix, after rotation, it appears in the penultimate column position:

The same is true for the third and fourth rows in the matrix.In this way we can get the law:

For the j-th element of the i-th row in the matrix, after rotation, it appears at the j-th position of the penultimate i-th column.

We translate it into code.Since the rows and columns in the matrix are counted from 000, for the element matrix[row][col] in the matrix, after the rotation, its new position is matrixnew[col][n−row−1].

In this way, we use an auxiliary array matrixnew{matrix} of the same size as matrix to temporarily store the rotated result.We traverse each element in the matrix and store the element in the corresponding position in matrixnew{matrix}_ according to the above rules.After the traversal is complete, copy the result in matrixnew{matrix}_ to the original array.

void rotate(int** matrix, int matrixSize, int* matrixColSize) {int matrix_new[matrixSize][matrixSize];for (int i = 0; i < matrixSize; i++) {for (int j = 0; j < matrixSize; j++) {matrix_new[i][j] = matrix[i][j];}}for (int i = 0; i < matrixSize; ++i) {for (int j = 0; j < matrixSize; ++j) {matrix[j][matrixSize - i - 1] = matrix_new[i][j];}}}