Java代碼??

/**?

?*?java?version?"1.6.0_17"<br>?

?*?尾遞歸與迭代實現具有相當的性能;<br>?

?*?緩存實現的性能略高于非尾遞歸實現;<br>?

?*?遞歸：recursive??尾遞歸：tail?recursive;<br>?

?*?尾遞歸時不需保存方法調用棧的參數及返回結果;于是申請的棧幀會被及時回收?

?*/??

public?class?TestFibo?{??

??

????public?static?void?main(String[]?args)?{??

????????int?N=50;??

????????long?begin=System.currentTimeMillis();????????

????????System.out.println(fibo1(N));?//fibo1??

????????System.out.println(System.currentTimeMillis()-begin);??

??????????

????????begin=System.currentTimeMillis();??

????????System.out.println(fibo2(N));?//fibo2??

????????System.out.println(System.currentTimeMillis()-begin);??

??????????

????????begin=System.currentTimeMillis();??

????????System.out.println(fibo3(N));?//fibo3??

????????System.out.println(System.currentTimeMillis()-begin);??

??????????

????????begin=System.currentTimeMillis();??

????????System.out.println(fibo4(N));?//fibo4??

????????System.out.println(System.currentTimeMillis()-begin);??

????}??

//1.1?非尾遞歸實現(書本上經常出現)????

public?static?long?fibo1(long?n){????

????if(n<2)?return?n;????

????return?fibo1(n-1)+fibo1(n-2);?//小心棧溢出????

}????

//1.2?緩存實現(JS?good?part里有過介紹)????

public?static?int?LENGTH=30;?//過大了就會占用過多空間????

public?static?long[]?cache=new?long[LENGTH];????

public?static?long?fibo2(int?n){????

????if(n<2)?return?n;????

????if(n>=LENGTH){????

????????return?fibo2(n-1)+fibo2(n-2);????

????}else?if(cache[n]==0){????

????????cache[n]=fibo2(n-1)+fibo2(n-2);?//減少重復計算????????????????

????}????

????return?cache[n];????

}????

//1.3?迭代實現????

public?static?long?fibo3(long?n){????

????if(n<2)?return?n;????

????long?pre=1,prepre=1,ret=0;????

????for(int?i=2;i<n;i++){????

????????ret=pre+prepre;????

????????prepre=pre;????

????????pre=ret;????

????}????

????return?ret;????

}????

//1.4?尾遞歸實現????

public?static?long?fibo4(int?n){????

????if(n<2)?return?n;????

????return?fibo4Helper(n,?1,?1,?3);?//保持與非尾遞歸接口不變，是借助幫助方法實現尾遞歸的????

}????

private?static?long?fibo4Helper(int?n,long?prepre,long?pre,int?begin){????

????if(n==begin)?return?pre+prepre;????

????return?fibo4Helper(n,?pre,?prepre+pre,?++begin);????//這里相當于迭代實現for-loop的濃縮??????

}???

??????

????//----------------------尾遞歸的其他實現-------------------------------------->??

????//2.?求最大公約數??

????public?static?int?gcd(int?big,int?small){??

????????if(big%small==0)?return?small;??

????????return?gcd(small,?big%small);??

????}??

????//3.1?階乘--非尾遞歸??

????public?static?int?fn1(int?n){??

????????if(n<2)?return?1;??

????????return?n*fn1(n-1);??

????}??

????//3.2?階乘--尾遞歸??

????public?static?int?fn2(int?n){??

????????if(n<2)?return?1;??

????????return?fn2Helper(1,?n);??

????}??

????private?static?int?fn2Helper(int?ret,?int?n){??

????????if(n<2)?return?ret;??

????????return?fn2Helper(ret*n,n-1);??

????}??

????????//4.1?翻轉字符串--非尾遞歸??

???????public?static?String?reverse1(String?s,?int?length){??

????????????if(length==0)?return?"";?//下一行的"+"可借助高版本JDK編譯器的優化??

????????????return?s.charAt(length-1)+reverse1(s,length-1);??

????????}??

???????//4.2?翻轉字符串--尾遞歸??

???????public?static?String?reverse2(String?s){??

???????????return?reverse2Helper(s,?"",?s.length());??

???????}??

???????private?static?String?reverse2Helper(String?s,String?init,int?len){??

???????????if(len==0)?return?init;??

???????????return?reverse2Helper(s,?init+s.charAt(len-1),?len-1);??

???????}??

????//5.?驗證字符串是否是回文?testHuiwen("abcdcba")??

?????public?static?boolean?testHuiwen(String?s){??

?????????return?testHuiwenHelper(s,?0,?s.length());??

?????}??

?????private?static?boolean?testHuiwenHelper(String?s,int?begin,?int?len){??

?????????if(begin<?len>>1?&&?s.charAt(begin)==s.charAt(len-begin-1))???

?????????????return?testHuiwenHelper(s,?begin+1,?len);??

?????????else?if(begin==len>>1)?return?true;??

?????????else?return?false;??

?????}??

????//6.?翻轉整數?如:1024=>4201??

?????public?static?int?reverseInt(int?i){??

?????????return?reverseIntHelper(i,?0);??

?????}??

?????private?static?int?reverseIntHelper(int?i,?int?init){??

?????????if(i==0)?return?init;??

?????????return?reverseIntHelper(i/10,?init*10+i%10);??

?????}??

}??

最后show一下強大的scala，用一行代碼獲取斐波那契數列的前30位：?
Scala代碼??

>(1?to?28).foldLeft("0,1")((a,b)=>a+","+a.split(",").takeRight(2).map(_.toInt).reduceLeft(_+_))????

??

>0,1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765,10946,17711,28657,46368,75025,121393,196418,317811,514229????

??

>(1?to?28).foldLeft(List(1,0))((a,b)=>(a.head+a.tail.head)::a).reverse.mkString(",")????

??

>0,1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765,10946,17711,28657,46368,75025,121393,196418,317811,514229??

??

??

scala>?def?fibo(n:Int)=(1?to?n).foldLeft(List(0,1))((a,b)=>a:+(a.init.last+a.last))??

fibo:?(n:?Int)List[Int]??

??

scala>?fibo(10)??

res4:?List[Int]?=?List(0,?1,?1,?2,?3,?5,?8,?13,?21,?34,?55,?89)??

總結

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