std::prev_permutation
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| definiert in Header <algorithm>
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| template< class BidirIt > bool prev_permutation( BidirIt first, BidirIt last); |
(1) | |
| template< class BidirIt, class Compare > bool prev_permutation( BidirIt first, BidirIt last, Compare comp); |
(2) | |
Transformiert den Bereich
[first, last) in die vorherige Permutation aus der Menge aller Permutationen, lexikographisch bezüglich operator< oder comp geordnet sind. Versandkosten true wenn solche Permutation existiert, sonst verwandelt den Bereich in den letzten Permutation (wie von std::sort(first, last); std::reverse(first, last);) und kehrt false .Original:
Transforms the range
[first, last) into the previous 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 last permutation (as if by std::sort(first, last); std::reverse(first, last);) and returns false.The text has been machine-translated via Google Translate.
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Inhaltsverzeichnis |
[Bearbeiten] Parameter
| first, last | - | das Spektrum der Elemente vertauschen
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: bool cmp(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. |
| Type requirements | ||
-BidirIt must meet the requirements of ValueSwappable and BidirectionalIterator.
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[Bearbeiten] Rückgabewert
true wenn die neue Permutation geht der alte in lexikographische Ordnung. false wenn die erste Permutation erreicht wurde und der Bereich wurde zurückgesetzt, um die letzte Permutation .
Original:
true if the new permutation precedes the old in lexicographical order. false if the first permutation was reached and the range was reset to the last permutation.
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[Bearbeiten] Komplexität
An den meisten
(last-first)/2 Swaps .Original:
At most
(last-first)/2 swaps.The text has been machine-translated via Google Translate.
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[Bearbeiten] Mögliche Implementierung
template<class BidirIt> bool prev_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 (*i1 < *--i) { i2 = last; while (!(*--i2 < *i)) ; std::iter_swap(i, i2); std::reverse(i1, last); return true; } if (i == first) { std::reverse(first, last); return false; } } } |
[Bearbeiten] Beispiel
Der folgende Code gibt alle sechs Permutationen der Zeichenfolge "abc" in umgekehrter Reihenfolge
Original:
The following code prints all six permutations of the string "abc" in reverse order
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#include <algorithm> #include <string> #include <iostream> #include <functional> int main() { std::string s="abc"; std::sort(s.begin(), s.end(), std::greater<char>()); do { std::cout << s << ' '; } while(std::prev_permutation(s.begin(), s.end())); std::cout << '\n'; }
Output:
cba cab bca bac acb abc
[Bearbeiten] Siehe auch
| (C++11) |
determines if a sequence is a permutation of another sequence (Funktions-Template) |
| prev_permutation |
generates the next smaller lexicographic permutation of a range of elements (Funktions-Template) |
