std::next_permutation
Defined in header <algorithm>
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template< class BidirIt > bool next_permutation( BidirIt first, BidirIt last ); |
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template< class BidirIt, class Compare > bool next_permutation( BidirIt first, BidirIt last, Compare comp ); |
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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.
Parameters
first, last | - | the range of elements to permute |
comp | - | comparison function object (i.e. an object that satisfies the requirements of Compare ) 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 object must not modify the objects passed to it. |
Type requirements | ||
-BidirIt must meet the requirements of ValueSwappable and BidirectionalIterator .
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Return value
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.
Exceptions
Any exceptions thrown from iterator operations or the element swap.
Complexity
At most N/2 swaps, where N = std::distance(first, last).
Possible implementation
template<class BidirIt> bool next_permutation(BidirIt first, BidirIt last) { if (first == last) return false; BidirIt i = last; if (first == --i) return false; while (true) { 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; } } } |
Example
The following code prints all three permutations of the string "aba"
#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
See also
(C++11) |
determines if a sequence is a permutation of another sequence (function template) |
generates the next smaller lexicographic permutation of a range of elements (function template) |