题解 | #跳台阶#
跳台阶
http://www.nowcoder.com/practice/8c82a5b80378478f9484d87d1c5f12a4
归纳法:
第一步:第一行是number值,第二行是步数,归纳完就是f(n) = f(n-1) + f(n-2)
第二步:找循环,number = 3,需要循环1次。number = 4,需要循环2次... ...
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# -*- coding:utf-8 -*- class Solution: def jumpFloor(self, number): # write code here if number <= 0 : return 0 elif number == 1: return 1 elif number == 2: return 2 else: a = 1 b = 2 i = 0 while i <= number-3: i+=1 a, b = b, a+b return b