LeetCode in Ruby (11 Part Series)
def climb_stairs(n) p = 1 q = 1 (n - 1).times do temp = q q += p p = temp end q end
This problem can be solved by using dynamic programming. We define f(n) as the number of ways you climb to the nth step from:
- one step from
- two steps from
f(n) = f(n-1) + f(n-2), which is the same as the Fibonacci sequence. We have two bases cases:
f(1) = 1 and
f(2) = 2. The lines 4 and 5 are the first two numbers of the Fibonacci sequence. To get the nth number, we add up the numbers up to nth number. Here, we optimize the performance the just storing the previous two numbers.
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