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]; } };
