Leetcode 867 - 转置矩阵
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2024-03-22 10:58:58
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题目
给定一个矩阵 A, 返回 A 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:[[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:[[1,2,3],[4,5,6]]
输出:[[1,4],[2,5],[3,6]]
提示:
- 1 <= A.length <= 1000
- 1 <= A[0].length <= 1000
分析
该题非常简单,理解转置矩阵的操作即可。对矩阵进行转置操作如下图所示
实现
C语言代码实现如下
/**
* 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** transpose(int** A, int ASize, int* AColSize, int* returnSize, int** returnColumnSizes){
int row = ASize;
int col = AColSize[0];
int **rslt = NULL;
int i, j;
int idx = 0;
/* 创建col行row列的二维数组 */
rslt = (int**)calloc(col, sizeof(int*));
for (i = 0; i < col; i++) {
rslt[i] = (int*)calloc(row, sizeof(int));
}
/* 矩阵转换 */
for (i = 0; i < row; i++) {
for (j = 0; j < col; j++) {
rslt[j][i] = A[i][j];
}
}
/* 转置矩阵的行数 */
*returnSize = col;
/* 转置矩阵每一行对应的列数 */
*returnColumnSizes = (int*)calloc(*returnSize, sizeof(int));
for (i = 0; i < *returnSize; i++) {
(*returnColumnSizes)[i] = row;
}
return rslt;
}
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