问题
I'm working on an algorithm to calculate a Fibonacci number and got the pseudo code for it but I can't figure out how much time it takes to run. I think it runs at O(n) but not quite sure. Here is the code:
Algorithm Fast-Fibonacci(n)
Let fib[0] and fib[1] be 1.
for each i from 2 to n, do:
Let fib[i] be fib[i - 2] + fib[i - 1].
end of loop
return fib[n].
Thanks for any help.
回答1:
You are correct that this takes O(n) as you are just counting sequentially from 2 to n to fill your array. If you were doing some sort of lookup for each of the i-1 and i-2 numbers, that could increase complexity, but the way you have written it, you are calling a direct address for each of those values.
回答2:
Yep. The big giveaway is that you have a constant number of operations per loop and the size of your loop is linear against the size of n.
A more space-efficient solution exists, however, since you don't particularly care about any numbers other than the last two. Try that next!
回答3:
Iterative solution to find the nth fibonnaci takes O(n) in terms of the value of n and O(2^length(n)) in terms of the size of n ( length(n) == number of bits to represent n). This kind of running time is called Pseudo-polynomial. However, if recursive method is used to find the fib of n, then it will take exponential time in terms of the value(O(2^n)). But this can be reduced by using dynamic programming approach to solve the fib of n.
来源：https://stackoverflow.com/questions/11475638/dynamic-programming-fibonacci-algorithm