도전

정수 목록이 주어지면 인접한 동일한 항목의 모든 쌍을 반복적으로 제거한 후 이러한 정수 목록을 반환하십시오.

홀수 길이의 같은 수의 런이 있으면 그 중 하나가 쌍의 일부가 아닌 그대로 유지됩니다.

예:

[0, 0, 0, 1, 2, 4, 4, 2, 1, 1, 0]

첫째, 당신은 제거해야합니다 0, 0, 4, 4그리고 1, 1얻을 :

[0, 1, 2, 2, 0]

이제 다음을 제거해야합니다 2, 2.

[0, 1, 0]

그리고 이것이 최종 결과입니다.

테스트 사례

[]-> []
[1]-> [1]
[1, 1]-> []
[1, 2]-> [1, 2]
[11, 11, 11]-> [11]
[1, 22, 1]-> [1, 22, 1]
[-31, 46, -31, 46]-> [-31, 46, -31, 46]
[1, 0, 0, 1]-> []
[5, 3, 10, 10, 5]-> [5, 3, 5]
[5, 3, 3, 3, 5]-> [5, 3, 5]
[0, -2, 4, 4, -2, 0]-> []
[0, 2, -14, -14, 2, 0, -1]-> [-1]
[0, 0, 0, 1, 2, 4, 4, 2, 1, 1, 0]-> [0, 1, 0]
[3, 5, 4, 4, 8, 26, 26, 8, 5]-> [3]
[-89, 89, -87, -8, 8, 88]-> [-89, 89, -87, -8, 8, 88]

채점

이것은 code-golf 이므로 각 언어에서 가장 짧은 답변이 이깁니다!

답변

Œgœ^/€FµÐL

Œgœ^/€FµÐL  Main link. Argument: A (array)

       µ    Combine all links to the left into a chain.
Œg              Group all adjacent equal items.
    /€          Reduce each group by...
  œ^                symmetric multiset difference.
                In each step, this maps ([], n) to [n] and ([n], n) to [], so the
                group is left with a single item if its length is odd, and no items
                at all if its length if even.
      F         Flatten the resulting array of singleton and empty arrays.
        ÐL  Apply the chain until the results are no longer unique. Return the last
            unique result.

+m`^(.+)¶\1$¶?

#//.{b___,a_,a_,c___}:>{b,c}&

r=[]
for x in input():r+=x,;r[-2:]*=r[-2:-1]!=[x]
print r

Œr;ṪḂ$$€x/€FµÐL

Œr;ṪḂ$$€x/€FµÐL  Main Link
Œr               Run-length encode
  ;              Concatenate (?)
       €         For each element
   ṪḂ$$          Is the last element odd?
          €      For each element    // Non-breaking alternative
        x/       Reduce by repeating // for run-length decode
           F     Flatten
            µ    (New monadic link)
             ÐL  Repeat until results are no longer unique

f=a=>1/a.find(q=>q==a[++i],i=-2)?f(a,a.splice(i,2)):a

r=[]                            # create list that will hold final results. A new list is important because it needs to be removable.
for i in input():
 if r!=[]and r[-1]==i:r.pop()   # Ensure that we have at least 1 char added to the list (r!=[])... or that the last character of our final result isn't the current character being scanned. If that is, well, remove it from the final list because we do not want it anymore
 else:r+=[i]                    # Shorthand for r.append(i). This adds i to the final result
print r