LeetCode 70 Climbing Stairs(Python详解及实现)

【题目】

You are climbing a stair case. It takes nsteps to reach to the top.

 

Each time you can either climb 1 or 2steps. In how many distinct ways can you climb to the top?

 

Note: Given n will be a positive integer.

 

爬楼梯，一次可以爬一步或者两步。爬n层，问一共有多少种。

【思路】DP

如果n = 1层，可以有[1] ,   f(1) = 1种方法

如果n = 2层，可以有[1,1] ,[2],  f(2) =  2种方法

如果n = 3层，可以有[1,1,1],[1,2],[2,1], f(3) = f(1) + f(2) = 3种方法。

如果n = 4层，可以有[1,1,1,1],[1,2,1],[1,1,2],[2,1,1],[2,2] , f(5)  = f(4) + f(3) = 5种方法。

可以发现 f(n) = f(n-1) + f(n-2)

【Python实现】

class Solution(object):

   def climbStairs(self, n):

       """

       :type n: int

       :rtype: int

       """

       if n == 1 or n == 0:

            return 1

       i,stairs1,stairs2 = 2,1,1

#i = 2 表示从2层楼梯开始，stairs1staris2 分别是n - 2, n-1 层楼梯的结果，

       while i <= n:

            stairs1 = stairs1 + stairs2

            if i == n:# n = 2 ，4时

                 print(stairs1)

                 return stairs1

            i += 1

            stairs2 = stairs1 + stairs2

            if i == n:# n = 3 ，5时

                 print(stairs2)

                 return stairs2

            i += 1

 

s = Solution()

s.climbStairs(4)