Leetcode 867. Transpose Matrix, Java解法

最新推荐文章于 2021-05-31 22:14:17 发布

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博客围绕矩阵转置展开，介绍了一种直接复制的暴力解法。给出了矩阵转置的定义，即矩阵沿主对角线翻转，交换行列索引。还进行了复杂度分析，时间复杂度和空间复杂度均为O(R * C)，同时提醒矩阵不一定是方的。

Given a matrix A, return the transpose of A.

The transpose of a matrix is the matrix flipped over it's main diagonal, switching the row and column indices of the matrix.

Example 1:

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

Example 2:

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

Note:

1 <= A.length <= 1000
1 <= A[0].length <= 1000

Approach #1 Brute Force, Copy Directly 我的解法

class Solution {
    public int[][] transpose(int[][] A) {
    int row = A.length;
    int col = A[0].length;
    int[][] ans = new int[col][row];
    for(int i=0;i<col;++i){
        for(int j=0;j<row;++j){
            ans[i][j] = A[j][i];
        }
    }
    return ans;
    }
}

Complexity Analysis

Time Complexity: O(R * C), where R and C are the number of rows and columns in the given matrix A.
Space Complexity: O(R * C), the space used by the answer.

Tips:

需要注意这个矩阵不一定是方的，审题要注意。