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;
    }
};

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Climbing Stairs

原文：http://blog.csdn.net/wan_hust/article/details/38307563