代码 Θ(nLogn) 的时间复杂度如何？

Question

 int n;
 int i, j, k = 0;
    for (i  = n/2; i <= n; i++) {
        for (j = 2; j <= n; j = j * 2) {
            k = k + n/2;
        }
    }

Just need to calculate the time complexity of the code snippet and the answer is Θ(nLogn) but can you explain how it is Θ(nLogn)

Answer 1

It's really not that difficult.

The outer loop runs n/2 times. That is O(n) complexity.

The inner loop again depends on n only. It doesn't depend on the i from the first loop. So for the complexity of the inner loop we can ignore completely the outer loop. How fortunate!. j multiples each time by 2 so we have logarithm base 2 . That is O(log(n)) .

The loops are nested so we multiply, thus ending with:

O(n log(n))