迭代算法的时间复杂度？

Question

This is the algorithm: I think its time complexity is O(nlogn) but i am not sure

k=1;
while (k<=n) do      
    j=1;
    while (j<=k) do        
        sum=sum+1;
        j=j+1;
    k=k*2;

Answer 1

The inner loop at the first time performs 1 iteration, at the second 2 iterations. The sequence goes like 1, 2, 4, 8, 16, 32, ... as long as it is smaller than or equal to n . The sequence may have Θ(log(n)) elements but its sum is Θ(n) . This is because

1 + 2 + 4 + ... + 2^k = 2 * 2^k - 1

and we know n/2 < 2^k <= n . So the inner loop is performed Θ(n) times and each inner loop execution requires constant number of instructions.

The rest of the code is just log(n) assignments to j = 1 and log(n) doubles of k .

So the time complexity of the algorithm is Θ(n) .

Answer 2

// code                 | max times executed
k=1;                    | 1
while (k<=n) do         | log n
    j=1;                | log n
    while (j<=k) do     | log n * n
        sum=sum+1;      | log n * n
        j=j+1;          | log n * n
    k=k*2;              | log n

So the O complexity would seem to be n log n.