斐波那契数列及简单dp应用

斐波那契数

• 问题：现在要求输入一个整数n，请你输出斐波那契数列的第n项（从0开始，第0项为0，第1项 是1）。 n<=39

• 解决：

  #方法一：迭代
  class Solution:
      def fib(self, N: int) -> int:
          if N < 2:
              return N
          first = 0
          second = 1
          third = 1
          while( N>=2):
              third = first + second
              first , second = second, third
              N -= 1
          return third
  #方法二：递归
  class Solution:
      def fib(self, N: int) -> int:
          if N < 2:
              return N
          return self.fib(N-1) + self.fib(N-2)
  #方法三：递归，剪枝减少重复计算
  class Solution:
      def fib(self, N: int) -> int:
          if N <= 1:
              return N
          # self.cache = {}
          # if N-1 in self.cache.keys():
          #     pre = self.cache[N-1]
          # else:
          #     pre = self.fib(N-1)
          #     self.cache[N-1] = pre
          # if N-2 in self.cache.keys():
          #     ppre = self.cache[N-2]
          # else:
          #     ppre = self.fib(N-2)
          #     self.cache[N-2] = ppre
          # return pre + ppre
          self.cache = {0:0,1:1}
          return self.filter(N)
      def filter(self, N: int) -> int:
          if N in self.cache.keys():
              return self.cache[N]
          self.cache[N] = self.filter(N-1)+self.filter(N-2)
          return self.filter(N)
  

• 问题：一只青蛙一次可以跳上1级台阶，也可以跳上2级。求该青蛙跳上一个n级的台阶总共有多少种跳法（先后次序不同算不同的结果）。

• 解决：

  #动态规划dp
  #1.定义状态：f(n):青蛙跳上第n级台阶的总跳法
  #2.转移方程：f(n) = f(n+1) + f(n+2)
  #3.设置初始值：f(0) = 1 , f(1) = 1, f(2) = 2
  class Solution:
      def jumpFloor(self, number):
          if number <= 1:
              return 1
          dp = {0:1, 1:1}
          for i in range(2,number+1):
              dp[i] = dp[i-1] + dp[i-2]
          return dp[number]