斐波那契數列定義：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]

本例以后一種為例：

最簡單的一種：兩層遞歸

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);}

問題是：隨著n的數值逐漸增多，時間和空間耗費太大，讀者可以自行實驗。在我的機器上n=50時就不能忍受了。

考慮優化：一層遞歸

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

遞歸時間減少了到不到50%

最好的方式，不使用遞歸的方式來做。

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;}

?

轉載于:https://www.cnblogs.com/davidwang456/p/4031167.html

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