题解 | 斐波那契数列

#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param n int整型 
# @return int整型
#
class Solution:
    def p(self, n):
        dp = [0] * (n+1)
        dp[0]=0
        dp[1] = 1
      
        for i in range(2, n+1):
            dp[i] = dp[i-1] + dp[i-2]
        return dp[n]
    def Fibonacci(self , n: int) -> int:
        # write code here
        return self.p(n)