70. Climbing Stairs

1. 问题描述

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?
Tags：Dynamic Programming

2. 解题思路

　　简单分析之：

f(1) = 1;

f(2) = 2;

f(3) = f(2) + f(1) = 3

...

f(n) = f(n-1) + f(n-2)

即，Fibonacci数列

3. 代码

class Solution
{
public:
//递归方法：在LeetCode测试，发现超时
int climbStairs_recur(int n)
{
if (1 == n)
{
return  1;
}
else if (2 == n)
{
return 2;
}
else
{
return climbStairs_recur(n-2) + climbStairs_recur(n-1);
}
}
//循环方法：利用了上一次的计算结果，速度快
int climbStairs_loop(int n)
{
if (1 == n)
{
return  1;
}
else if (2 == n)
{
return 2;
}

int cur = 2;
int last = 1;
for (int i=3; i<=n; i++)
{
int t =  last + cur;
last = cur;
cur = t;
}
return cur;
}
};

4. 反思