leetcode专题训练 70. Climbing Stairs

很简单的一道dp。当前状态可能由上一块跳1步或上上块跳2步得到。所以 $dp[i]=dp[i-1]+dp[i-2]$
需要注意 $n==1$的特殊情况

class Solution:
    def climbStairs(self, n: int) -> int:
        if n == 1:
            return 1
        dp = [1 for i in range(n)]
        dp[1] = 2
        for i in range(2, n):
            dp[i] = dp[i-1]+dp[i-2]
        return dp[-1]