[勇者闯LeetCode] 70. Climbing Stairs
[勇者闯LeetCode] 70. Climbing Stairs
Description
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.
Information
- Tags: Dynamic Programming
- Difficulty: Easy
Solution
实质上是Fibonacci Number：
Fib(n)=Fib(n−1)+Fib(n−2),Fib(1)=1,Fib(2)=2 Fib(n)=Fib(n−1)+Fib(n−2), Fib(1)=1, Fib(2)=2，
即爬到n阶的方法数等于爬到n-1阶的方法数和爬到n-2阶的方法数之和，爬到1阶的方法数是1，爬到2阶的方法数是2。
class Solution(object):def climbStairs(self, n):""":type n: int:rtype: int"""prev, cur = 0, 1for i in range(n):prev, cur = cur, cur + prevreturn cur[勇者闯LeetCode] 70. Climbing Stairs相关推荐
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