题目描写叙述：

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?

思路:斐波那契数列的应用。

f(n)=f(n-1)+f(n-2)。当中f(1)=1,f(2)=2。此题假设用递归会超时，所以採用非递归的方法求斐波那契数列的第n项。

代码：

int Solution::climbStairs(int n){    int n1 = 1;    int n2 = 2;    if(n == 1)        return 1;    if(n == 2)        return 2;    while(n > 2)    {        int temp = n1;        n1 = n2;        n2 = n2 + temp;        n--;    }    return n2;}