[LeetCode OJ]-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?

一共有n阶楼梯，每次只能爬1阶或2阶，问爬到顶端一共有多少种方法

方法一：

利用二叉树的思想，对于n=3时有3种方法，有几个叶子节点便有几种方法

 1 void climb( int remainder, int &way)
 2 {
 3     if(remainder==0 || remainder==1)
 4     {
 5         way++;
 6         return;
 7     }
 8 
 9     climb(remainder-1, way);
10     climb(remainder-2, way);
11     return;
12 }
13 
14 class Solution {
15 public:
16     int climbStairs(int n) {
17         int result=0;
18         climb(n, result);
19         return result;
20     }
21 };

但是这种方法对于n比较大时会超时，在测试用例中对于n=38就会TLE（Time Limit Exceed）。

方法二：

总结规律，通过以下数据，我们发现

way(1)=1

way(n) = way(n-1) + Fibonacci(n-1) n=2,3,4,5,6,....

 1 class Solution {
 2 public:
 3     int climbStairs(int n) {
 4         int result=1;
 5         int Fibonacci_0=0, Fibonacci_1=1, temp;
 6         for(int i=1; i<n; i++)
 7         {
 8             result += Fibonacci_1;
 9             temp = Fibonacci_1;
10             Fibonacci_1 = Fibonacci_0 + Fibonacci_1;
11             Fibonacci_0 = temp;
12 
13         }
14         return result;
15     }
16 };

对于leetcode中所有的测试数据都可以通过