斐波那契数列（fabnacci）java实现

斐波那契数列定义：From Wikipedia, the free encyclopedia

http://en.wikipedia.org/wiki/Fibonacci_number

In mathematics, the Fibonacci numbers or Fibonacci sequence are the numbers in the following integer sequence:^[2][3]

or (often, in modern usage):

 (sequence A000045 in OEIS).

By definition, the first two numbers in the Fibonacci sequence are 1 and 1, or 0 and 1, depending on the chosen starting point of the sequence, and each subsequent number is the sum of the previous two.

In mathematical terms, the sequence F_n of Fibonacci numbers is defined by the recurrence relation

with seed values^[2][3]

or^[4]

本例以后一种为例：

最简单的一种：两层递归

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

问题是：随着n的数值逐渐增多，时间和空间耗费太大，读者可以自行实验。在我的机器上n=50时就不能忍受了。

考虑优化：一层递归

    public static void main(String[] args) {
        long tmp=0;
        // TODO Auto-generated method stub
        int n=10;
        Long start=System.currentTimeMillis();
        for(int i=0;i<n;i++){
            System.out.print(fibonacci(i)+" ");
        }
        System.out.println("-------------------------");
        System.out.println("耗时："+(System.currentTimeMillis()-start));
    }    

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

递归时间减少了到不到50%

最好的方式，不使用递归的方式来做。

    public static long fibonacci(int n){
        long before=0,behind=0;
        long result=0;
        for(int i=0;i<n;i++){
            if(i==0){
                result=0;
                before=0;
                behind=0;
            }
            else if(i==1){
                result=1;
                before=0;
                behind=result;
            }else{
                result=before+behind;
                before=behind;
                behind=result;

            }
        }
        return result;
    }