递归可能超时，所以使用循环来做。 思想和递归相同。从3开始往后计算跳法。

-- coding:utf-8 --

class Solution: def jumpFloor(self, number): # write code here # write code here t1, t2, total = 1,2,0 i = 3 if number <=2: return number

    while i <= number:
        total = t1 + t2
        t1 = t2
        t2 = total
        i += 1
    return total