class Solution:
    # def __init__(self):
    #         self.dic = {1:1, 2:2}

    def climbStairs(self, n):
        """
        :type n: int
        :rtype: int
        """
        # 這道題本質上是一個斐波那契數列,因為後一個數總是等於前兩個數字之和!

        # 思路一:  Top down - TLE  因為前面已經算過的數字不能直接呼叫,總是需要重新計算.
        # if n == 1:
 

        #     return 1
        # if n == 2:
        #     return 2
        # if n > 2:
        #     return self.climbStairs(n-1) + self.climbStairs(n-2)

        # Top down + memorization (list) 對於思路一的改進
#         def climbStairs4(self, n):
#             if n == 1:
#                 return 1
#             dic = [-1 for i in xrange(n)]
 

#             dic[0], dic[1] = 1, 2
#             return self.helper(n-1, dic)

#         def helper(self, n, dic):
#             if dic[n] < 0:
#                 dic[n] = self.helper(n-1, dic)+self.helper(n-2, dic)
#             return dic[n]

        # Top down + memorization (dictionary)  對於思路一的改進,用一個字典來儲存前面計算過的部分,減少運算時間
 

        # if n not in self.dic:
        #     self.dic[n] = self.climbStairs(n-1) + self.climbStairs(n-2)
        # return self.dic[n]




        # Bottom up, O(n) space  思路二:自底向上. 儲存前面計算過的數字,用空間來換取時間
        # if n == 1:
        #     return 1
        # res = [0 for i in range(n)]
        # res[0], res[1] = 1, 2
        # for i in range(2, n):
        #     res[i] = res[i-1] + res[i-2]
        # return res[-1]

        # Bottom up, O(n) space  思路二:自底向上. 進一步壓縮佔用的儲存空間
        if n == 1:
            return 1
        a, b = 1, 2
        for i in range(2, n):
            a,b = b,a+b
        return b