leetcode-climbing stairs

第一种是迭代，第二种是DP，第三种是斐波那契数列的通项公式。

此外，soulmachine的书上还指出，递归也可以，只是速度太慢。

#include <iostream>
#include <cmath>
using namespace std;

class Solution {
public:
    int climbStairs(int n) {
        int prev = 0;
        int cur = 1;
        for (int i = 1; i <= n; i++)
        {
            int tmp = cur;
            cur += prev;
            prev = tmp;
        }
        return cur;
    }
};
class Solution2 {
public:
    int climbStairs(int n) {
        int* dp = new int[n + 1];
        dp[1] = 1; dp[0] = 1;
        for (int i = 2; i <= n; i++)
        {
            dp[i] = dp[i - 1] + dp[i - 2];
        }
        return dp[n];
    }
};
class Solution3 {//attention: we should calculate a[n+1] instead of a[n]!!!because a[1]=1,a[2]=1!!!
public:
    int climbStairs(int n) {
        int result;
        double s = sqrt(5);
        result = (1 / s)*(pow((1 + s) / 2, n + 1) - pow((1 - s) / 2, n + 1));
        return result;
    }
};

int main()
{
    //Solution s;
    //Solution2 s;
    Solution3 s;
    cout << s.climbStairs(8) << endl;
    return 0;
}

https://github.com/soulmachine/leetcode

http://rleetcode.blogspot.com/2014/06/climbing-stairs.html