LeetCode#70 Climbing Stairs

Problem Definition:

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?

 

Solution: 可以用动态规划来解。令爬n级阶梯的方式有f(n)种。

　　要到达第n级阶梯，要么是从第n-1级阶梯爬一步，要么是从第n-2级阶梯爬两步（一次性）。而这两种情况是没有交集的。

　　因此得到：f(n)=f(n-1)+f(n-2)  (当n>1)

　　oops...正是斐波那契数列。

def climbStairs(n):
    a,b,i=[1]*3
    while i<n:
        a,b=b,a+b
        i+=1
    return b