Python爬楼梯，一次只能上一阶或者两阶，一共有n阶，计算一共有多少种上法，用动态规则算法

Question

Python爬楼梯，一次只能上一阶或者两阶，一共有n阶，计算一共有多少种上法，用动态规则算法，求问

Answer 1

n = 3
class Solution(object):
    def climbStairs(self, n):
        """
        :type n: int
        :rtype: int
        """
        # base case
        if n == 0:
            return(0)
        if n == 1:
            return(1)
        if n == 2:
            return(2)
        
       
        methods_per_steps = [1,2]
        
        for i in range(2,n+1):
            methods_per_steps.append(methods_per_steps[i-1]+methods_per_steps[i-2])
        return(methods_per_steps[n-1])
        
out = Solution().climbStairs(n)
print(out)