快速斐波那契递归 [英] Fast Fibonacci recursion

问题描述

我试图回忆起在斐波那契递归算法。以下内容：

I'm trying to recall an algorithm on Fibonacci recursion. The following:

public int fibonacci(int n)  {
  if(n == 0)
    return 0;
  else if(n == 1)
    return 1;
  else
    return fibonacci(n - 1) + fibonacci(n - 2);
}

是不可以我在找什么，因为它是贪婪的。这将成倍增长（只是看 Java的递归Fibonacci序列 - 越大的初始参数更没用的电话将进行）。

is not what I'm looking for because it's greedy. This will grow exponentially (just look at Java recursive Fibonacci sequence - the bigger the initial argument the more useless calls will be made).

有可能是有点像循环论证移位，其中调用previous斐波那契价​​值将再次检索计算它的价值吧。

There is probably something like a "cyclic argument shift", where calling previous Fibonacci value will retrieve value instead of calculating it again.

推荐答案

也许是这样的：

int fib(int term, int val = 1, int prev = 0)
{
 if(term == 0) return prev;
 if(term == 1) return val;
 return fib(term - 1, val+prev, val);
}

此功能是尾递归。这意味着它可以优化和执行非常有效。事实上，它被优化成一个简单的循环。

this function is tail recursive. this means it could be optimized and executed very efficiently. In fact, it gets optimized into a simple loop..

这篇关于快速斐波那契递归的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！