bool next_permutation( BidirIt first, BidirIt last, Compare comp ); (2) (C++20 起为 constexpr) 将范围 [first, last) 变换为下个排列。这种排列存在时返回 true,否则将范围变换为首个排列(如同用 std::sort)并返回 false。 1) 所有排列的集合按相对于 operator<(C++20 前)std::less{}(C++20 起...
}while(next_permutation(s.begin(), s.end()));return0; } 这里的next_permutation的实现思想跟第二部分中全排列的非递归实现的思想一样. 下面给出cppreference.com 提供的一种next_permutation的实现方法: template<classBidirIt>boolnext_permutation(BidirIt first, BidirIt last){if(first == last)returnf...
Result = next_permutation ( deq1.begin ( ) , deq1.end ( ) ); if ( deq1Result ) cout << "The lexicographically next permutation " << "exists and has\nreplaced the original " << "ordering of the sequence in deq1." << endl; else cout << "The lexicographically next permutation ...
After another next_permutation of vector v1, v1 = ( -3 -2 -1 0 3 2 1 ). After another next_permutation of vector v1, v1 = ( -3 -2 -1 1 0 2 3 ).RequirementsHeader: <algorithm>Namespace: stdSee AlsoReferencenext_permutation (STL Samples)...
template<classBidirIt>boolnext_permutation(BidirIt first, BidirIt last){autor_first=std::make_reverse_iterator(last);autor_last=std::make_reverse_iterator(first);autoleft=std::is_sorted_until(r_first, r_last);if(left!=r_last){autoright=std::upper_bound(r_first, left,*left);std::iter...
Result = next_permutation ( deq1.begin ( ) , deq1.end ( ) ); if ( deq1Result ) cout << "The lexicographically next permutation " << "exists and has\nreplaced the original " << "ordering of the sequence in deq1." << endl; else cout << "The lexicographically next permutation ...
The default binary predicate is less than and the elements in the range must be less than comparable to insure that the next permutation is well defined. The complexity is linear with at most (_Last – _First)/2 swaps. Example コピー // alg_next_perm.cpp // compile with: /EHsc #...
Result = next_permutation ( deq1.begin ( ) , deq1.end ( ) ); if ( deq1Result ) cout << "The lexicographically next permutation " << "exists and has\nreplaced the original " << "ordering of the sequence in deq1." << endl; else cout << "The lexicographically next permutation ...
next_permutation, vector v1 is: v1 = ( -3 -2 -1 0 1 3 2 ). After another next_permutation of vector v1, v1 = ( -3 -2 -1 0 2 1 3 ). After another next_permutation of vector v1, v1 = ( -3 -2 -1 0 2 3 1 ). After another next_permutation of vector v1, v1...