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) | |
變換的範圍內
[first, last)進入下一置換從該組的所有字典順序排列的排列相對於operator<或comp。返回true如果這樣的排列存在，否則將到第一個置換（如果由std::sort(first, last)），並返回false.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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目錄 |
[编辑] 參數
| first, last | - | 元素的範圍內的重排
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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[编辑] 返回值
true如果新的排列是大於舊字典。 false如果最後的排列，達到的範圍複位到第一排列.
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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[编辑] 複雜性
在最N/2掉期，N = std::distance(first, last).
Original:
At most N/2 swaps, where N = std::distance(first, last).
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[编辑] 可能的實現
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; } } } |
[编辑] 為例
下面的代碼列印「ABA」字元串的所有排列
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
[编辑] 另請參閱
| （C++11） |
determines if a sequence is a permutation of another sequence （函數模板） |
| generates the next smaller lexicographic permutation of a range of elements （函數模板） | |
