题解 | #跳台阶#

#coding:utf-8
#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param number int整型 
# @return int整型
#
class Solution:
    def jumpFloor(self , n):
        # write code here
        #debug
        ##dp[i] 代表到第i台阶有多少种办法
        #dp[0] = 0
        #dp[1] = 1
        #dp[2] = 2 = dp[1] + dp[0] = dp[1] + dp[2 - 2]
        #dp[n] = dp[n - 1] + dp[n - 2]
        #init
        dp = [0] * (n + 1)
        dp[0] = 1
        dp[1] = 1
        
        #process
        for i in range(2, n + 1):
            dp[i] = dp[i - 2] + dp[i - 1]
        #print ("dp: ", dp)

        return dp[n]