你不需要在Haskell中使用memoize函数。只有在面向过程的编程语言中才需要这些函数。然而,Haskell是一种函数式语言,所以...
因此,以下是非常快速的斐波那契算法示例:
fib = zipWith (+) (0:(1:fib)) (1:fib)
zipWith是来自标准Prelude的函数:
zipWith :: (a->b->c) -> [a]->[b]->[c]
zipWith op (n1:val1) (n2:val2) = (n1 + n2) : (zipWith op val1 val2)
zipWith _ _ _ = []
测试:
print $ take 100 fib
输出:
[1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765,10946,17711,28657,46368,75025,121393,196418,317811,514229,832040,1346269,2178309,3524578,5702887,9227465,14930352,24157817,39088169,63245986,102334155,165580141,267914296,433494437,701408733,1134903170,1836311903,2971215073,4807526976,7778742049,12586269025,20365011074,32951280099,53316291173,86267571272,139583862445,225851433717,365435296162,591286729879,956722026041,1548008755920,2504730781961,4052739537881,6557470319842,10610209857723,17167680177565,27777890035288,44945570212853,72723460248141,117669030460994,190392490709135,308061521170129,498454011879264,806515533049393,1304969544928657,2111485077978050,3416454622906707,5527939700884757,8944394323791464,14472334024676221,23416728348467685,37889062373143906,61305790721611591,99194853094755497,160500643816367088,259695496911122585,420196140727489673,679891637638612258,1100087778366101931,1779979416004714189,2880067194370816120,4660046610375530309,7540113804746346429,12200160415121876738,19740274219868223167,31940434634990099905,51680708854858323072,83621143489848422977,135301852344706746049,218922995834555169026,354224848179261915075,573147844013817084101]
经过的时间:0.00018秒
fib 0
没有终止:你可能希望fib'
的基本情况是fib' 0 = 0
和fib' 1 = 1
。 - huonfibs = 1:1:zipWith (+) fibs (tail fibs)
和fib = (fibs !!)
。 - Bastian