std::is_permutation
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Defined in header <algorithm> | ||
template<class ForwardIt1, class ForwardIt2 > bool is_permutation( ForwardIt1 first, ForwardIt1 last, | (1) | (C++11およびそれ以降) |
template<class ForwardIt1, class ForwardIt2, class BinaryPredicate > bool is_permutation( ForwardIt1 first, ForwardIt1 last, | (2) | (C++11およびそれ以降) |
戻りtrue
[first1, last1)で範囲の先頭に、その範囲が等しくなる範囲d_first内の要素の順列が存在する場合。最初のバージョンは、平等のためにoperator==を使用して、2番目のバージョンは、バイナリ述語pを使用していますOriginal:
Returns true if there exists a permutation of the elements in the range
[first1, last1) that makes that range equal to the range beginning at d_first. The first version uses operator== for equality, the second version uses the binary predicate pThe text has been machine-translated via Google Translate.
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目次 |
[編集]パラメータ
| first, last | - | 比較するための要素の範囲 Original: the range of elements to compare The text has been machine-translated via Google Translate. You can help to correct and verify the translation. Click here for instructions. |
| d_first | - | 第2の範囲の始まりは、比較することができます Original: the beginning of the second range to compare The text has been machine-translated via Google Translate. You can help to correct and verify the translation. Click here for instructions. |
| p | - | binary predicate which returns true if the elements should be treated as equal. The signature of the predicate function should be equivalent to the following: bool pred(const Type1 &a, const Type2 &b); The signature does not need to have const&, but the function must not modify the objects passed to it. |
| 型の要件 | ||
-ForwardIt1, ForwardIt2 は ForwardIteratorの要求を満足しなければなりません。 | ||
[編集]値を返します
true範囲
[first, last)はd_firstで範囲の先頭の順列である場合.Original:
true if the range
[first, last) is a permutation of the range beginning at d_first.The text has been machine-translated via Google Translate.
You can help to correct and verify the translation. Click here for instructions.
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[編集]複雑性
述部の中で最もO(N2)アプリケーションで、または正確にN配列が既に等しい場合、どこN=std::distance(first, last).
Original:
At most O(N2) applications of the predicate, or exactly N if the sequences are already equal, where N=std::distance(first, last).
The text has been machine-translated via Google Translate.
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[編集]可能な実装
template<class ForwardIt1, class ForwardIt2>bool is_permutation(ForwardIt1 first, ForwardIt1 last, ForwardIt2 d_first){// skip common prefixstd::tie(first, d_first)=std::mismatch(first, last, d_first);// iterate over the rest, counting how many times each element// from [first, last) appears in [d_first, d_last)if(first != last){ ForwardIt2 d_last = d_first;std::advance(d_last, std::distance(first, last));for(ForwardIt1 i = first; i != last;++i){if(i !=std::find(first, i, *i))continue;// already counted this *i auto m =std::count(d_first, d_last, *i);if(m==0||std::count(i, last, *i)!= m){returnfalse;}}}returntrue;} |
[編集]例
このコードを実行します
#include <algorithm>#include <vector>#include <iostream>int main(){std::vector<int> v1{1,2,3,4,5};std::vector<int> v2{3,5,4,1,2};std::cout<<"3,5,4,1,2 is a permutation of 1,2,3,4,5? "<<std::boolalpha<< std::is_permutation(v1.begin(), v1.end(), v2.begin())<<'\n'; std::vector<int> v3{3,5,4,1,1};std::cout<<"3,5,4,1,1 is a permutation of 1,2,3,4,5? "<<std::boolalpha<< std::is_permutation(v1.begin(), v1.end(), v3.begin())<<'\n';}
出力:
3,5,4,1,2 is a permutation of 1,2,3,4,5? true 3,5,4,1,1 is a permutation of 1,2,3,4,5? false
[編集]参照
| generates the next greater lexicographic permutation of a range of elements (関数テンプレート) | |
| generates the next smaller lexicographic permutation of a range of elements (関数テンプレート) | |
