18.5 Iterative Fibonacci

Iterative Fibonacci

# iteration is much quicker than recursion for this problem

# finds the nth term in the Fibonacci sequence using iteration
def iterative_fib(n):
    if n <= 0:
        return None  # base case; out of bounds

    current = 0
    next_term = 1

    for i in range(n - 1):  # this is equivalent to for i in range(1, n)
        current, next_term = next_term, current + next_term
        # this is just a slightly rewritten fib sequence;
        # instead of looking at the past 2 cases, it looks at the
        # current and next terms to determine the next next term

    return current  # will be 0 if n is 1, 1 if n is 2, etc...

print(iterative_fib(5))