[LeetCode]Climbing Stairs

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?
求斐波那契， f(n) = f(n-1)+f(n-2)

f(1) = 1; f(2) = 2;

1.递归,超慢

class Solution {
public:
    int help(int n)
    {
		if(n==1)
			return 1;
		if(n==2)
			return 2;
		return help(n-1)+help(n-2);
	}
	int climbStairs(int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        return help(n);
    }
	
};

2.迭代

class Solution {
public:
    int climbStairs(int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int ret;
		if(n==1)
			return 1;
		if(n==2)
			return 2;
		int x = 2;
		int y = 1;
        int z = 0;
		int i = 2;
		while( i<n)
		{
			z = x + y;
            i ++ ;
			y = x;
			x = z;
		}
		return z;
    }
	
};

3.矩阵公式法

A= [1 1]

[1 0]

A^n = [F(n) , F(n-1)]

[F(n-1),F(n-2)]

关键是 n为 偶数的 时候 拆分成 (A^(2\n)) ^2计算 奇数 就再乘以一个A

运算速度更快，写的不太好，可能要压栈耗费很多空间。。。

class Solution {
public:
    vector<int> pow(int n){
        if(n==1){
    		vector<int> tmp;  
			tmp.push_back(1);
			tmp.push_back(1);
			tmp.push_back(1);
			tmp.push_back(0);
			return tmp;
		}
			vector<int> tmp,tmp1;
            tmp.clear();
            tmp1.clear();
            tmp.resize(4);
            tmp1.resize(4);
			tmp =pow(n>>1);
			tmp1[0] = tmp[0]* tmp[0] + tmp[1] * tmp[2];
			tmp1[1] = tmp[0]* tmp[1] + tmp[1] * tmp[3];
			tmp1[2] = tmp1[1];
			tmp1[3] = tmp[1]* tmp[2] + tmp[3] * tmp[3];	
		if(n%2)
		{
			tmp[0] = tmp1[0] + tmp1[1];
			tmp[1] = tmp1[0] ;
			tmp[2] = tmp1[0] ;
			tmp[3] = tmp1[1] ;
			return tmp;
		}
		return tmp1;
			
	}
    int climbStairs(int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<int> ret;
        ret.resize(4);
        ret =pow(n);
        return ret[0];
    }
	
};

4.无理数公式（有精度问题）

略