【算法作业11】LeetCode 70. Climbing Stairs

70. 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?

Note: Given n will be a positive integer.

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题解：

这也是一道动态规划的题目，但仔细一看其实它的本质就是斐波那契数列。因为每次只能走一步或者两步，也就是说如果一共有n个台阶的话，就相当于是在n-1个台阶的基础上再走一步，或者在n-2个台阶的基础上走两步，所以有n个台阶的方法总数就是有n-2个台阶和有n-1个台阶的方法之和。刚开始用了递归的斐波那契算法，运行结果是正确的但是超时了，所以后面就改成了用数组实现斐波那契。

代码：

class Solution {public:    int climbStairs(int n) {        int * fib = new int[n + 1];        fib[0] = 0;        fib[1] = 1;        fib[2] = 2;        for (int i = 3; i <= n; i++)        {            fib[i] = fib[i - 1] + fib[i - 2];        }        return fib[n];    }};