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std::is_permutation

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< cpp‎ | algorithm

 
 
アルゴリズムライブラリ
機能します
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Functions
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シーケンス動作を非改変
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Non-modifying sequence operations
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シーケンス動作を変更する
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Modifying sequence operations
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操作を仕切る
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Partitioning operations
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(ソートされた範囲で)ソート操作
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Sorting operations (on sorted ranges)
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バイナリ検索操作(ソート範囲で)
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Binary search operations (on sorted ranges)
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(ソートされた範囲で)操作を設定します
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Set operations (on sorted ranges)
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ヒープ操作
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Heap operations
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最小値/最大値操作
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Minimum/maximum operations
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数値演算
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Numeric operations
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Cライブラリ
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C library
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Defined in header <algorithm>
template< class ForwardIt1, class ForwardIt2 >

bool is_permutation( ForwardIt1 first, ForwardIt1 last,

                     ForwardIt2 d_first );
(1) (C + + 11以来)
template< class ForwardIt1, class ForwardIt2, class BinaryPredicate >

bool is_permutation( ForwardIt1 first, ForwardIt1 last,

                     ForwardIt2 d_first, BinaryPredicate p );
(2) (C + + 11以来)
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 p
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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
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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
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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.
The types  Type1 and  Type2 must be such that objects of types ForwardIt1 and ForwardIt2 can be dereferenced and then implicitly converted to  Type1 and  Type2 respectively.

Type requirements
-
ForwardIt1, ForwardIt2 must meet the requirements of ForwardIterator.

[編集] 値を返します

true範囲[first, last)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.
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[編集] 複雑

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).
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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 prefix
   std::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) {
                return false;
            }
        }
    }
    return true;
}

[編集]

#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';
}

Output:

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] も参照してください

generates the next greater lexicographic permutation of a range of elements
(関数テンプレート) [edit]
generates the next smaller lexicographic permutation of a range of elements
(関数テンプレート) [edit]