急求！！！！python列表里面根据一定的条件挑选元素各位有什么建议？

Question

我有一个list如下结构：L=["ticket1","ticket2","spec1","spec2","ticket3","ticket4","spec3","ticket5","spec4","spec5","ticket6","ticket7","ticket8","ticket9","ticket10","spec6","spec7","ticket11","ticket12","ticket13","spec8","ticket14","spec9","ticket15","ticket16","ticket17","ticket18","spec1""ticket19","ticket20","spec2""ticket21"]
想要挑出来后面没有跟着spec的ticket，具体规则如下：1.如果连续两个ticket后面跟着连续两个spec，那么这两个ticket算有spec，如ticket1，ticket2;ticket9,ticket102.如果一个ticket后面跟着一个或者连续多个spec,那么这个ticket也算有spec，如ticket4,ticket5,ticket13,ticket14;
剩下的ticket都算是没有spec的，我想要挑出这些ticket，有哪些好的方法可以使用呀？没有spec的ticket如下：ticket3,ticket6,ticket7,ticket8,ticket11,ticket12,ticket15,ticket16,ticket17,ticket19,ticket21
列表里面的信息是唯一的信息，请教各位大神有什么方法可以实现这样的挑选。

Answer 1

应该是2*n的时间复杂度，可以按这个原理再优化成n的时间复杂度和O(1)的空间复杂度L=["ticket1","ticket2","spec1","spec2","ticket3","ticket4","spec3","ticket5","spec4","spec5","ticket6","ticket7","ticket8","ticket9","ticket10","spec6","spec7","ticket11","ticket12","ticket13","spec8","ticket14","spec9"]defcalc(l):iflen(l)=3:ifs>=2:returnl[:t-2]else:returnl[:t-s]elift>s:returnl[:t-s]return[]defmain(L):res=[]start=0fori,curr_eleinenumerate(L[:-1]):next_ele=L[i+1]ifcurr_ele.startswith('spec')andnext_ele.startswith('ticket'):res.extend(calc(L[start:i+1]))start=i+1res.extend(calc(L[start:]))returnresprint(main(L))['ticket3','ticket6','ticket7','ticket8','ticket11','ticket12']

Answer 2

update:之前一版是错的，忽略了两层栈深还必须ticket、spce连续的要求换个解法，代码有些冗长#!/usr/bin/envpython#-*-coding:utf-8-*-defis_ticket(node):returnnode.startswith('ticket')defis_spec(node):returnnode.startswith('spec')defdeal1(L):ifL:node=L.pop(0)#无论何种，都会使表长-1ifis_ticket(node):returnnodereturnNonedefdeal2(L):defmatch_ts(L):node1,node2=L[:2]returnis_ticket(node1)andis_spec(node2)iflen(L)find_no_spect(L)['ticket6','ticket7','ticket8','ticket11','ticket12']