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prev_permutation
Permutes the order of the element
Key Facts
Gyroscopic Couple: The rate of change of angular momentum (= Moment of Inertia.
= Angular velocity
= Angular velocity of precession.
Blaise Pascal (1623-1662) was a French mathematician, physicist, inventor, writer and Catholic philosopher.
Leonhard Euler (1707-1783) was a pioneering Swiss mathematician and physicist.
Definition
The prev_permutation() algorithm is defined in the standard header <algorithm> and in the nonstandard backward-compatibility header <algo.h>.Interface
#include <algorithm> template < class BidirectionalIterator > bool prev_permutation( BidirectionalIterator first, BidirectionalIterator last ); template < class BidirectionalIterator, class BinaryPredicate > bool prev_permutation( BidirectionalIterator first, BidirectionalIterator last, BinaryPredicate comp );Parameters:
| Parameter | Description |
|---|---|
| first | A bidirectional iterator pointing to the position of the first element in the range to be permuted |
| last | A bidirectional iterator pointing to the position one past the final element in the range to be permuted |
| comp | User-defined predicate function object that defines the comparison criterion to be satisfied by successive elements in the ordering. A binary predicate takes two arguments and returns true when satisfied and false when not satisfied |
Description
Prev_permutation function reorders the range[first, last) into the previous permutation from the set of all permutations that are lexicographically ordered.
The first version uses operator< for comparison and the second uses the function object comp.Return Value
Returnstrue if there was a previous permutation and has replaced the original ordering of the range, false otherwise.Example:
Example - prev_permutation algorithm
Problem
The following code prints all six permutations of the string "abc" in reverse order using prev_permutation function.
Workings
#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'; }
Solution
Output:
cba cab bca bac acb abc
References