C#LeetCode刷题之#867-转置矩阵（Transpose Matrix）

问题

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。

输入：[[1,2,3],[4,5,6],[7,8,9]]

输出：[[1,4,7],[2,5,8],[3,6,9]]

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

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.

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

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

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

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

Note:

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

示例

public class Program {

    public static void Main(string[] args) {
        int[][] nums = null;

        nums = new int[3][] {
            new int[] {1, 2, 3},
            new int[] {4, 5, 6},
            new int[] {7, 8, 9}
        };

        var res = Transpose(nums);
        ShowArray(res);

        Console.ReadKey();
    }

    private static void ShowArray(int[][] array) {
        foreach(var line in array) {
            foreach(var num in line) {
                Console.Write($"{num} ");
            }
        }
        Console.WriteLine();
    }

    private static int[][] Transpose(int[][] A) {
        var m = A.Length;
        var n = A[0].Length;
        var result = new int[n][];
        for(int i = 0; i < n; i++) {
            result[i] = new int[m];
        }
        for(var i = 0; i < n; i++) {
            for(var j = 0; j < m; j++) {
                result[i][j] = A[j][i];
            }
        }
        return result;
    }

}

以上给出1种算法实现，以下是这个案例的输出结果：

1 4 7 2 5 8 3 6 9

分析：

显而易见，以上算法的时间复杂度为： 。