题目
文本
给你一个二维整数数组 matrix, 返回 matrix 的 转置矩阵 。 矩阵的 转置 是指将矩阵的主对角线翻转,交换矩阵的行索引与列索引。
示例 1:
输入:matrix = [[1,2,3],[4,5,6],[7,8,9]] 输出:[[1,4,7],[2,5,8],[3,6,9]]
示例 2:
输入:matrix = [[1,2,3],[4,5,6]] 输出:[[1,4],[2,5],[3,6]]
提示:
m == matrix.length
n == matrix[i].length
1 <= m, n <= 1000
1 <= m * n <= 105
-109 <= matrix[i][j] <= 109
来源:力扣(LeetCode)
模板
/**
* 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) {
}
解题
分析
这题不难就是对角线互换元素即可 这里我加上了数组下标,就更能发现转化前后的关系。 很明显看出就是二维数组的下标进行交换。
- 二维数组初始化,行列互换
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;
}
- 元素填充
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
transposed[j][i] = matrix[i][j];
}
}
就这么简单。
完整源码
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;
}