[LeetCode] Climbing Stairs

最新推荐文章于 2019-05-12 12:35:00 发布

1.02^365的成长裂变

最新推荐文章于 2019-05-12 12:35:00 发布

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本文探讨了一个经典的编程面试题——爬楼梯问题。通过对问题进行数学抽象，将其转化为求解特定斐波那契数列的过程，并提供了一种高效的非递归算法实现方案。

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[Problem]

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?

[Analysis]
Fibnocci Sequency. We can climb n-2 stairs end with 2 stairs or climb n-1 stairs end with 1 stair to reach n stairs. That means f(n) = f(n-2) + f(n-1).

In case of TLE for a large n, we should not use a recursive style for this program.

[Solution]

class Solution {
public:
    int climbStairs(int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int fab[n+1];
        fab[0] = 1;
        fab[1] = 1;
        for(int i = 2; i <= n; ++i){
            fab[i] = fab[i-1] + fab[i-2];
        }
        return fab[n];
    }
};