我是这门ERLANG的新手，我知道一些基础知识.这就像计划，但范围更广.我知道如何创建函数，但是创建具有最长公共子序列的函数时遇到问题.
I'm new to this ERLANG, I know the basics. It's like scheme but broader. I know how to create a function but I'm having problems with creating a function that gets the Longest Common Subsequence.
lcs(str1,str2)是一个将接受两个字符串并输出一个整数的函数:
lcs(str1,str2) is a function that will accept two string and output an integer:
lcs(algorithm,logarithm)将输出7，因为可以进行的最长公共子序列是lgrithm，其大小为7.
lcs(algorithm,logarithm) will output 7 because the longest common subsequence that can be made is lgrithm which is 7 in size.
任何回答都将不胜感激.
Any answer is greatly appreciated.
推荐答案您在 Rosettacode ，即:
-module(lcs). -compile(export_all). lcs_length(S,T) -> {L,_C} = lcs_length(S,T,dict:new()), L. lcs_length([]=S,T,Cache) -> {0,dict:store({S,T},0,Cache)}; lcs_length(S,[]=T,Cache) -> {0,dict:store({S,T},0,Cache)}; lcs_length([H|ST]=S,[H|TT]=T,Cache) -> {L,C} = lcs_length(ST,TT,Cache), {L+1,dict:store({S,T},L+1,C)}; lcs_length([_SH|ST]=S,[_TH|TT]=T,Cache) -> case dict:is_key({S,T},Cache) of true -> {dict:fetch({S,T},Cache),Cache}; false -> {L1,C1} = lcs_length(S,TT,Cache), {L2,C2} = lcs_length(ST,T,C1), L = lists:max([L1,L2]), {L,dict:store({S,T},L,C2)} end. lcs(S,T) -> {_,C} = lcs_length(S,T,dict:new()), lcs(S,T,C,[]). lcs([],_,_,Acc) -> lists:reverse(Acc); lcs(_,[],_,Acc) -> lists:reverse(Acc); lcs([H|ST],[H|TT],Cache,Acc) -> lcs(ST,TT,Cache,[H|Acc]); lcs([_SH|ST]=S,[_TH|TT]=T,Cache,Acc) -> case dict:fetch({S,TT},Cache) > dict:fetch({ST,T},Cache) of true -> lcs(S,TT,Cache, Acc); false -> lcs(ST,T,Cache,Acc) end.用作:
raringbeast:Code pierre $ erl Erlang R15B01 (erts-5.9.1) [source] [64-bit] [smp:8:8] [async-threads:0] [hipe] [kernel-poll:false] Eshell V5.9.1 (abort with ^G) 1> c(lcs). {ok,lcs} 2> lcs:lcs("logarithm", "algorithm"). "lgrithm" 3> lcs:lcs_length("logarithm", "algorithm"). 7-
使算法更易于理解
在某些情况下，这里的缓存只是提高算法性能的一种有趣方式，但这并不是必需的.我们可以写，删除缓存:
The cache here is just an interesting way to improve algorithm performance in some cases, but this isn't required here. We can just write, removing the cache:
lcs_length([],_T) -> 0; lcs_length(_S,[]) -> 0; lcs_length([_H|ST],[_H|TT]) -> 1 + lcs_length(ST,TT); lcs_length([_SH|ST]=S,[_TH|TT]=T) -> L1 = lcs_length(S,TT), L2 = lcs_length(ST,T), lists:max([L1,L2]).总结: