Likou 867. Transpose Matrix-C Language Implementation-Simple Questions

topic

Portal

text

Gives you a two-dimensional integer array matrix, and returns the transposed matrix of the matrix.
The transposition of a matrix refers to flipping the main diagonal of the matrix and exchanging the row index and column index of the matrix.

Example 1:

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

Example 2:

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

prompt:

m == matrix.length
n == matrix[i].length
1 <= m, n <= 1000
1 <= m * n <= 105
-109 <= matrix[i][j] <= 109

Source: LeetCode

template

/**
 * Return an array of arrays of size *returnSize.
 * The sizes of the arrays are returned as *returnColumnSizes array.
 * Note: Both returned array and *columnSizes array must be malloced, assume caller calls free().
 *返回大小为* returnSize的数组。
 *数组的大小作为* returnColumnSizes数组返回。
 *注意：返回的数组和* columnSizes数组都必须被分配，假设调用者调用free（）。
 */
int** transpose(int** matrix, int matrixSize, int* matrixColSize, int* returnSize, int** returnColumnSizes) {
    
    

}

Problem solving

analysis

This question is not difficult, just swap the elements diagonally.

Here I added the array subscripts to better discover the relationship before and after conversion.
It is obvious that the subscripts of the two-dimensional array are exchanged.

• Two-dimensional array initialization, row and column swap

    int m = matrixSize, n = matrixColSize[0];
    int** transposed = malloc(sizeof(int*) * n);
    *returnSize = n;
    *returnColumnSizes = malloc(sizeof(int) * n);
    for (int i = 0; i < n; i++) {
    
    
        transposed[i] = malloc(sizeof(int) * m);
        (*returnColumnSizes)[i] = m;
    }

• Element fill

    for (int i = 0; i < m; i++) {
    
    
        for (int j = 0; j < n; j++) {
    
    
            transposed[j][i] = matrix[i][j];
        }
    }

It's that simple.

Complete source code

int** transpose(int** matrix, int matrixSize, int* matrixColSize, int* returnSize, int** returnColumnSizes) {
    
    
    int m = matrixSize, n = matrixColSize[0];
    int** transposed = malloc(sizeof(int*) * n);
    *returnSize = n;
    *returnColumnSizes = malloc(sizeof(int) * n);
    for (int i = 0; i < n; i++) {
    
    
        transposed[i] = malloc(sizeof(int) * m);
        (*returnColumnSizes)[i] = m;
    }
    for (int i = 0; i < m; i++) {
    
    
        for (int j = 0; j < n; j++) {
    
    
            transposed[j][i] = matrix[i][j];
        }
    }
    return transposed;
}

operation result