斐波那契数列(fabnacci)java实现
斐波那契數(shù)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):
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?Fn?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(jiǎn)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(shù)值逐漸增多,時(shí)間和空間耗費(fèi)太大,讀者可以自行實(shí)驗(yàn)。在我的機(jī)器上n=50時(shí)就不能忍受了。
考慮優(yōu)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(shí):"+(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(shí)間減少了到不到50%
最好的方式,不使用遞歸的方式來(lái)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(zhuǎn)載于:https://www.cnblogs.com/davidwang456/p/4031167.html
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