Leetcode:Climbing Stairs 斐波那契数

戳我去解题

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?

解析：设直至第n层有f(n)中走法，因为每次只能走1步或2步，则 f(n) = f(n-1) + f(n-2)

很明显是一个斐波那契数，f(0) = 0, f(1) = 1

1. 递归，  f(n) = f(n-1) + f(n-2) 很容易想到递归，但是递归过程中存在大量的重复计算，会超时

2. 迭代

class Solution {

public:

    int climbStairs(int n) {

        assert(n >= 0);

        int prevTwo = 0;

        int prevOne = 1;

        int cur = 1;

        for (int i = 1; i <= n; ++i) {

            prevOne = cur;

            cur = prevOne + prevTwo;

            prevTwo = prevOne;

        }

        return cur;

    }

};

3. 直接计算：

斐波那契直接可以计算出表达数出来，且计算方法有好几种：等比数列法、矩阵法、生成函数法

具体见：斐波那契解法

class Solution {

public:

    int climbStairs(int n) {

        assert(n >= 0);

        double s = sqrt(5);

        return floor((pow((1+s)/2, n+1) + pow((1-s)/2, n+1)) / s + 0.5);

    }

};