std::prev_permutation
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Defined in header <algorithm> | ||
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) | |
Transforme la gamme
[first, last) dans la permutation précédente de l'ensemble de toutes les permutations qui sont lexicographiquement ordonnées par rapport à operator< ou comp. true Retours s'il en existe permutation, autrement transforme la plage dans la dernière permutation (comme par std::sort(first, last); std::reverse(first, last);) et retourne 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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Sommaire |
[modifier]Paramètres
| first, last | - | la plage d'éléments à permuter 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. | ||
[modifier]Retourne la valeur
true si la nouvelle permutation précède l'ancien dans l'ordre lexicographique. false si la première permutation a été atteint et la gamme a été remis à la permutation dernière .
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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[modifier]Complexité
À la plupart des swaps
(last-first)/2 .Original:
At most
(last-first)/2 swaps.The text has been machine-translated via Google Translate.
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[modifier]Mise en œuvre possible
template<class BidirIt>bool prev_permutation(BidirIt first, BidirIt last){if(first == last)returnfalse; BidirIt i = last;if(first ==--i)returnfalse; while(1){ BidirIt i1, i2; i1 = i;if(*i1 <*--i){ i2 = last;while(!(*--i2 <*i));std::iter_swap(i, i2);std::reverse(i1, last);returntrue;}if(i == first){std::reverse(first, last);returnfalse;}}} |
[modifier]Exemple
Le code suivant imprime tous les six permutations de la chaîne "abc" dans l'ordre inverse
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';}
Résultat :
cba cab bca bac acb abc
[modifier]Voir aussi
(C++11) | determines if a sequence is a permutation of another sequence (fonction générique) |
prev_permutation | lexicographique génère le plus petit côté d'une permutation série d'éléments Original: generates the next smaller lexicographic permutation of a range of elements The text has been machine-translated via Google Translate. You can help to correct and verify the translation. Click here for instructions. (fonction générique) |
