python - 使用一个递归调用实现递归

Question

给定一个函数如下:f(n) = f(n-1) + f(n-3) + f(n-4)

f(0) = 1
f(1) = 2
f(2) = 3
f(3) = 4

我知道在一个函数内使用递归和三个递归调用来实现它。但我只想在函数内部进行一次递归调用。如何做到？

这里使用 3 个递归调用来实现是我的代码:

def recur(n):
  if n == 0:
     return 1
  elif n == 1:
     return 2
  elif n == 2:
     return 3
  elif n == 3:
     return 4
  else:
     return recur(n-1) + recur(n-3) + recur(n-4) #this breaks the rule because there are 3 calls to recur

Answer 1

您的尝试方向是正确的，但需要稍作改动:

def main():
  while True:
    n = input("Enter number : ")
    recur(1,2,3,4,1,int(n))

def recur(firstNum,secondNum,thirdNum,fourthNum,counter,n):  
  if counter==n:
     print (firstNum)
     return
  elif counter < n:
      recur (secondNum,thirdNum,fourthNum,firstNum+secondNum+fourthNum,counter+1,n)