leetcode-Climbing Stairs

题目描述：

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?

做了半天，就是运行不起来，最后根据结果输出，百度了一下这个数字，发现居然是斐波那契数列中的数，然后根据规律一看，答案果然是斐波那契数列，开始想复杂了，坑爹

上开始写的代码，数超过了int的最大范围，运行不起来

class Solution {public:    static int fun(int a,int b)    {        if(b==0)            return 1;        long long bigNum=a;        int index=1;        for(int i=1;i<b;i++)        {            bigNum=bigNum*(a-i);            index=index*(i+1);        }        return bigNum/index;    }    int climbStairs(int n) {        int index = n%2;        int num = n/2;        int sum = 0;        for(int i=0;i<=num;i++)        {            if((i*2+index)<(num-i))            {                sum=sum+fun(i*2+index+num-i,index+i*2);            }            else            {                sum=sum+fun(i*2+index+num-i,num-i);            }        }        return sum;    }};

为什么是斐波那契数列呢，原因是这样的，如果有n个台阶，那么就等于n-1个台阶走一步或是n-2个台阶走两步，所以是斐波那契数列

class Solution {public:    int climbStairs(int n) {        if(n<=2)            return n;        else        {            int *step=new int[n];            step[0]=1;            step[1]=2;            for(int i=2;i<n;i++)            {                step[i]=step[i-1]+step[i-2];            }            return step[n-1];        }    }};