std::next_permutation
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| Defined in header <algorithm>
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| template< class BidirIt > bool next_permutation( BidirIt first, BidirIt last ); |
(1) | |
| template< class BidirIt, class Compare > bool next_permutation( BidirIt first, BidirIt last, Compare comp ); |
(2) | |
辞書的に
[first, last)またはoperator<に対して順序付けされているすべての順列の集合から次の順列に範囲compを変換します。このような順列存在する場合は返品trueは、そうでなければ第一std::sort(first, last)順列（としてfalseであれば）を返しに範囲を変換.Original:
Transforms the range
[first, last) into the next permutation from the set of all permutations that are lexicographically ordered with respect to operator< or comp. Returns true if such permutation exists, otherwise transforms the range into the first permutation (as if by std::sort(first, last)) and returns false.The text has been machine-translated via Google Translate.
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目次 |
[編集] パラメータ
| first, last | - | ランダムな置換をする要素の範囲
Original: the range of elements to permute The text has been machine-translated via Google Translate. You can help to correct and verify the translation. Click here for instructions. | |||||||||
| comp | - | comparison function which returns true if the first argument is less than the second. The signature of the comparison function should be equivalent to the following:
The signature does not need to have const &, but the function must not modify the objects passed to it. | |||||||||
| Type requirements | |||||||||||
-BidirIt must meet the requirements of ValueSwappable and BidirectionalIterator.
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[編集] 値を返します
true新しい順列が古いより辞書式に大きい場合。 false最後の順列に達すると範囲は、最初の順列にリセットされていた場合..
Original:
true if the new permutation is lexicographically greater than the old. false if the last permutation was reached and the range was reset to the first permutation.
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[編集] 複雑
せいぜいN/2スワップ、どこN = std::distance(first, last).
Original:
At most N/2 swaps, where N = std::distance(first, last).
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[編集] 可能な実装
template<class BidirIt> bool next_permutation(BidirIt first, BidirIt last) { if (first == last) return false; BidirIt i = last; if (first == --i) return false; while (1) { BidirIt i1, i2; i1 = i; if (*--i < *i1) { i2 = last; while (!(*i < *--i2)) ; std::iter_swap(i, i2); std::reverse(i1, last); return true; } if (i == first) { std::reverse(first, last); return false; } } } |
[編集] 例
次のコードは、文字列 "ABA"の3つのすべての順列を出力します
Original:
The following code prints all three permutations of the string "aba"
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#include <algorithm> #include <string> #include <iostream> int main() { std::string s = "aba"; std::sort(s.begin(), s.end()); do { std::cout << s << '\n'; } while(std::next_permutation(s.begin(), s.end())); }
Output:
aab aba baa
[編集] も参照してください
| (C++11) |
determines if a sequence is a permutation of another sequence (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) | |
