LeetCode||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.

Example 1:

Input: 2
Output:  2
Explanation:  There are two ways to climb to the top.

1. 1 step + 1 step
2. 2 steps

Example 2:

Input: 3
Output:  3
Explanation:  There are three ways to climb to the top.

1. 1 step + 1 step + 1 step
2. 1 step + 2 steps
3. 2 steps + 1 step

爬楼梯问题。经典的动态规划问题。每次上一个台阶或者两个台阶，问一共有多少种方法到楼顶。这个实际上就是斐波那契数列的求解。可以逆向来分析问题，如果有n个台阶，那么走完n个台阶的方式有f(n)种。而走完n个台阶有两种方法，先走完n-2个台阶，然后跨2个台阶；先走完n-1个台阶，然后跨1个台阶。所以f(n) = f(n-1) + f(n-2)。
class Solution(object):
def climbStairs(self, n):
"""
:type n: int
:rtype: int
"""
dp = [1 for i in range(n+1)]
for i in range(2,n+1):
dp[i] = dp[i-1] + dp[i-2]
return dp