第十一周：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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我把前五项列出来，假如n=1那么有1种方法；假如n=2，那么有2种方法；假如n=3，那么有3种方法；假如n=4，那么有5种方法；假如n=5，那么有8种方法...这样列下去发觉这是一个斐波那契数列，用递归完成。

发现当用递归完成之后，会超时。

于是上网看了一下什么情况，原来发现递归解决这道题会出现超时错误，于是只能一步一步做出来。

Accepted Code：

int climbStairs(int n) { if(n == 0 || n == 1) return 1; int res = 0; int a = 1; int b = 1; for(int i = 2; i <= n; i ++) { res = a + b; a = b; b = res; } return res; }