Climbing Stairs

问题描述：

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?

解题思路：

这是一个典型的斐波那契数列，对于有n步的台阶：

1、如果先跳1步，还剩余n-1步，那么这就变成n-1步台阶跳法数目；

2、如果先跳2步，还剩余n-2步，这就变成n-2步台阶跳法数目。

所以经典的表达式：f(n) = f(n-1) + f(n-2)

class Solution {public:    int climbStairs(int n) {        /*f(n)=f(n-1)+f(n-2)*/        if(n==0||n==1) return  1;        int stepOne=1,stepTwo=1;        int result=0;        for(int i=2;i<=n;i++){            result=stepOne+stepTwo;            stepTwo=stepOne;            stepOne=result;        }        return result;    }};