Matlab中转置操作的Big O时间复杂度及两类场景咨询

Great question! Let's break this down clearly for both 1D and 2D array scenarios, since MATLAB handles these cases differently under the hood.

Scenario 1: 1D Array Transpose #

Suppose you define a 1D row vector:

a = [1 2 3 4];
b = a';

The time complexity here is O(1).

Here's why: MATLAB treats 1D arrays as row vectors by default, but transposing it to a column vector doesn't require copying any of the actual element data. Instead, MATLAB just updates the array's internal metadata (like changing the dimension tag from 1x4 to 4x1). There's no iteration over elements or memory copying involved in the transpose operation itself.

Quick note: If you later modify b, MATLAB might trigger a copy (thanks to its copy-on-write mechanism), but that's a separate operation—the transpose itself is constant time.

Scenario 2: 2D Matrix Transpose #

For a 2D matrix like this:

a = [ [1 2 3 4] ; [5 6 7 8] ; [9 10 11 12]]; % 3x4 matrix
b = a';

The time complexity here is O(m*n), where m is the number of rows and n is the number of columns in the original matrix.

The reason is MATLAB's column-major memory storage: elements are stored in memory column-by-column (so for the example above, the order is 1, 5, 9, 2, 6, 10, ...). When you transpose the matrix, the resulting 4x3 matrix needs its elements arranged in a completely different order to maintain column-major storage for the transposed structure. This requires copying every single element to a new memory location, hence linear time relative to the total number of elements in the matrix.

内容的提问来源于stack exchange，提问作者Mohamed Mohsen