Leetcode_num13_Climbing Stairs

标签：leetcode   台阶   

题目：

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

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

一开始准备用递归做，代码如下：

class Solution:
    # @param n, an integer
    # @return an integer
    def climbStairs(self, n):
        if n<=2:
            return n
        else:
            return self.climbStairs(n-1)+self.climbStairs(n-2)

于是，转向迭代了，代码如下：

class Solution:
    # @param n, an integer
    # @return an integer
    def climbStairs(self, n):
        if n<=1:
            return n
        else:
            s=[0 for i in range(n)]
            s[0]=1  #到达第1级
            s[1]=2  #到达第2级
            for i in range(2,n):
                s[i]=s[i-1]+s[i-2]
            return s[n-1] #到达第n级

于是进行改进，设s1为走一步到达方法数，s2为走两步到达的方法数。那么到达第n级台阶时，s(n)=s1+s2,其中s1=s(n-1),s2=s(n-2);到达第n+1级台阶时，s(n+1)=s1+s2,其中s1=s(n)=上一步的s1+s2, s2=s(n-1)=上一步的s1，所以只需要记录s1和s2的值，无需记录n个值

class Solution:
    # @param n, an integer
    # @return an integer
    def climbStairs(self, n):
        if n<=1:
            return n
        else:
            s1=1
            s2=1
            for i in range(1,n):
                s=s1+s2
                s2=s1
                s1=s
            return s 

Leetcode_num13_Climbing Stairs

标签：leetcode   台阶   

原文地址：http://blog.csdn.net/eliza1130/article/details/39646815