LeetCode:Climbing Stairs(编程之美2.9-斐波那契数列)

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?

算法1：分析：dp[i]为爬到第i个台阶需要的步数，那么dp[i] = dp[i-1] + dp[i-2], 很容易看出来这是斐波那契数列的公式本文地址

class Solution {
public:
    int climbStairs(int n) {
        int fbn1 = 0, fbn2 = 1;
        for(int i = 1; i <= n; i++)
        {
            int tmp = fbn1 + fbn2;
            fbn1 = fbn2;
            fbn2 = tmp;
        }
        return fbn2;
    }
};

算法2：还可以根据斐波那契数列的通项公式来求，对于斐波那契数列 1 1 2 3 5 8 13 21，通项公式如下，这个方法有个缺陷是：使用了浮点数，但是浮点数精度有限， oj中n应该不大，所以可以通过（当 N>93 时第N个数的值超过64位无符号整数可表示的范围）

具体推导请参考百度百科

class Solution {
public:
    int climbStairs(int n) {
        //根据斐波那契数列的通项公式
        double a = 1/sqrt(5);
        double b = (1 + sqrt(5)) / 2;
        double c = (1 - sqrt(5)) / 2;
        return (int)round(a * (pow(b, n+1) - pow(c, n+1)));
    }
};

算法3：”编程之美2.9-斐波那契数列“ 中提到了一种logn的算法（实际上利用了幂运算的logn算法），在n比较大时，会高效很多。首先给出本题代码，然后直接截图书上的描述。如果n较大，就需要编写大整数类了

 1     struct matrix22
 2     {
 3         int v11,v12,v21,v22;
 4         matrix22(int a,int b,int c,int d)
 5         {
 6             v11 = a; v12 = b; v21 = c; v22 = d;
 7         }
 8         matrix22(){}
 9     };
10     matrix22 matMult(const matrix22 &a, const matrix22 &b)//矩阵乘法
11     {
12         matrix22 res;
13         res.v11 = a.v11*b.v11 + a.v12*b.v21;
14         res.v12 = a.v11*b.v12 + a.v12*b.v22;
15         res.v21 = a.v21*b.v11 + a.v22*b.v21;
16         res.v22 = a.v21*b.v12 + a.v22*b.v22;
17         return res;
18     }
19     matrix22 matPow(const matrix22 &a, int exp)//矩阵求幂
20     {
21         matrix22 res(1,0,0,1);//初始化结果为单位矩阵
22         matrix22 tmp = a;
23         for(; exp; exp >>= 1)
24         {
25             if(exp & 1)
26                 res = matMult(res, tmp);
27             tmp = matMult(tmp, tmp);
28         }
29         return res;
30     }
31 
32 class Solution {
33 public:
34     int climbStairs(int n) {
35         matrix22 A(1,1,1,0);
36         A = matPow(A, n-1);
37         return A.v11 + A.v21;
38     }
39 };
40 
41