https://leetcode.cn/problems/transpose-matrix/description/
创建一个二级指针模版，只需要将数组中的i和j变换位置即可；

/*** 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().*/int **myMalloc(int r, int c, int * returnSize, int ** returnColumnSizes) {int ** ret = (int **)malloc(sizeof(int *) * r);*returnColumnSizes = (int *)malloc(sizeof(int) * r);*returnSize = r;for(int i = 0; i < r; ++i) {ret[i] = (int *)malloc(sizeof(int) * c);(*returnColumnSizes)[i] = c;}return ret;
}int** transpose(int** matrix, int matrixSize, int* matrixColSize, int* returnSize, int** returnColumnSizes) {int r = matrixColSize[0];int c = matrixSize;int ** res = myMalloc(r, c, returnSize, returnColumnSizes);for(int i = 0; i < r; ++i) {for(int j = 0; j < c; ++j) {res[i][j] = matrix[j][i];}}return res;
}

https://leetcode.cn/problems/flipping-an-image/description/

/*** 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().*/int ** myMalloc(int r, int c, int * returnSize, int ** returnColumnSizes) {int ** ret = (int **)malloc(sizeof(int *) * r);*returnSize = r;*returnColumnSizes = (int *)malloc(sizeof(int) * r);for(int i = 0; i < r; ++i) {ret[i] = (int *)malloc(sizeof(int) * c);(*returnColumnSizes)[i] = c;}return ret;
}int** flipAndInvertImage(int** image, int imageSize, int* imageColSize, int* returnSize, int** returnColumnSizes) {int n = imageSize;int ** res = myMalloc(n, n, returnSize, returnColumnSizes);for(int i = 0; i < n; ++i) {for(int j = 0; j < n; ++j) {res[i][j] = 1 - image[i][n - 1 - j];}}return res;}

https://leetcode.cn/problems/convert-1d-array-into-2d-array/description/
第一种:
这种方法好理解，创建二维数组判断是否能够构成二维数组都不难；
如果m * n != 一维数组长度则不能构成;
一维变为二维那里，可以先遍历原数组，然后根据二维数组每一行的开始元素都等于一维数组的0 n 2n 3n...判断是否可以整除n

int **myMalloc(int r, int c, int * returnSize, int **returnColumnSizes) {int **ret = (int **)malloc(sizeof(int *) * r);*returnSize = r;*returnColumnSizes = (int *)malloc(sizeof(int) * r);for(int i = 0; i < r; ++i) {ret[i] = (int *)malloc(sizeof(int) * c);(*returnColumnSizes)[i] = c;}return ret;
}int** construct2DArray(int* original, int originalSize, int m, int n, int* returnSize, int** returnColumnSizes) {int ** res = myMalloc(m, n, returnSize, returnColumnSizes);if(m * n != originalSize) {*returnSize = 0;return NULL;}int tr = 0, tc = 1;res[0][0] = original[0];for(int i = 1; i < originalSize; ++i) {if(i % n == 0) {tr++;tc = 0;}res[tr][tc++] = original[i];}return res;
}

https://leetcode.cn/problems/shift-2d-grid/description/

/*** 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().*/int **myMalloc(int r, int c, int * returnSize, int ** returnColumnSizes) {// 创建返回二级指针数组int **ret = (int **)malloc(sizeof(int *) * r);// returnColumnSizes是二级指针,加一个*是一级指针,代表行;*returnColumnSizes = (int *)malloc(sizeof(int) * r);// returnSize解引用是常量,代表有多少行;*returnSize = r;// 遍历ret,设置每行列for(int i = 0; i < r; ++i) {ret[i] = (int *)malloc(sizeof(int) * c);(*returnColumnSizes)[i] = c;}return ret;
}int** shiftGrid(int** grid, int gridSize, int* gridColSize, int k, int* returnSize, int** returnColumnSizes){int r = gridSize;int c = gridColSize[0];int ** res = myMalloc(r, c, returnSize, returnColumnSizes);for(int i = 0; i < r; ++i) {for(int j = 0; j < c; ++j) {int index = (i * c + j + k) % (r * c);res[index / c][index % c] = grid[i][j];}}return res;
}

https://leetcode.cn/problems/image-smoother/description/

/*** 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().*/int **myMalloc(int r, int c, int * returnSize, int ** returnColumnSizes) {// 创建返回二级指针数组int **ret = (int **)malloc(sizeof(int *) * r);// returnColumnSizes是二级指针,加一个*是一级指针,代表行;*returnColumnSizes = (int *)malloc(sizeof(int) * r);// returnSize解引用是常量,代表有多少行;*returnSize = r;// 遍历ret,设置每行列for(int i = 0; i < r; ++i) {ret[i] = (int *)malloc(sizeof(int) * c);(*returnColumnSizes)[i] = c;}return ret;
}
int rn[8] = {-1, -1, -1, 0, 0, 1, 1, 1};
int cn[8] = {-1, 0, 1, -1, 1, -1, 0, 1};
int** imageSmoother(int** img, int imgSize, int* imgColSize, int* returnSize, int** returnColumnSizes) {int r = imgSize;int c = imgColSize[0];int **res = myMalloc(r, c, returnSize, returnColumnSizes);for(int i = 0; i < r; ++i) {for(int j = 0; j < c; ++j) {int temp = img[i][j];int s = 1;for(int x = 0; x < 8; ++x) {int ir = i + rn[x];int ic = j + cn[x];if(ir < 0 || ic < 0 || ir >= r || ic >= c) {continue;}else {temp += img[ir][ic];s++;}}temp /= s;res[i][j] = temp;}}return res;
}

https://leetcode.cn/problems/matrix-block-sum/description/
这道题正规解法应该是前缀和，但是由于我才疏学浅，只能用四重循环暴力解出来了；
没想到竟然还能过~~~

 int **myMalloc(int r, int c, int * returnSize, int ** returnColumnSizes) {// 创建返回二级指针数组int **ret = (int **)malloc(sizeof(int *) * r);// returnColumnSizes是二级指针,加一个*是一级指针,代表行;*returnColumnSizes = (int *)malloc(sizeof(int) * r);// returnSize解引用是常量,代表有多少行;*returnSize = r;// 遍历ret,设置每行列for(int i = 0; i < r; ++i) {ret[i] = (int *)malloc(sizeof(int) * c);(*returnColumnSizes)[i] = c;}return ret;
}int** matrixBlockSum(int** mat, int matSize, int* matColSize, int k, int* returnSize, int** returnColumnSizes) {int m = matSize;int n = matColSize[0];int ** ret = myMalloc(m, n, returnSize, returnColumnSizes);for(int i = 0; i < m; ++i) {for(int j = 0; j < n; ++j) {ret[i][j] = 0;for(int x = i - k; x <= i + k; ++x) {if(x < 0 || x >= m)continue;for(int y = j - k; y <= j + k; ++y) {if(y < 0 || y >= n) continue;ret[i][j] += mat[x][y];}}}}return ret;
}