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is_sorted

The header file <boost/algorithm/cxx11/is_sorted.hpp> contains functions for determining if a sequence is ordered.

is_sorted

The function is_sorted(sequence) determines whether or not a sequence is completely sorted according so some criteria. If no comparison predicate is specified, then std::less_equal is used (i.e, the test is to see if the sequence is non-decreasing)

namespace boost { namespace algorithm {
	template <typename ForwardIterator, typename Pred>
	bool is_sorted ( ForwardIterator first, ForwardIterator last, Pred p );
	
	template <typename ForwardIterator>
	bool is_sorted ( ForwardIterator first, ForwardIterator last );
	
	
	template <typename Range, typename Pred>
	bool is_sorted ( const Range &r, Pred p );
	
	template <typename Range>
	bool is_sorted ( const Range &r );
}}

Iterator requirements: The is_sorted functions will work forward iterators or better.

is_sorted_until

If distance(first, last) < 2, then is_sorted ( first, last ) returns last. Otherwise, it returns the last iterator i in [first,last] for which the range [first,i) is sorted.

In short, it returns the element in the sequence that is "out of order". If the entire sequence is sorted (according to the predicate), then it will return last.

namespace boost { namespace algorithm {
	template <typename ForwardIterator, typename Pred>
	FI is_sorted_until ( ForwardIterator first, ForwardIterator last, Pred p );
	
	template <typename ForwardIterator>
	ForwardIterator is_sorted_until ( ForwardIterator first, ForwardIterator last );
	
	
	template <typename Range, typename Pred>
	typename boost::range_iterator<const R>::type is_sorted_until ( const Range &r, Pred p );
	
	template <typename Range>
	typename boost::range_iterator<const R>::type is_sorted_until ( const Range &r );
}}

Iterator requirements: The is_sorted_until functions will work on forward iterators or better. Since they have to return a place in the input sequence, input iterators will not suffice.

Complexity: is_sorted_until will make at most N-1 calls to the predicate (given a sequence of length N).

Examples:

Given the sequence { 1, 2, 3, 4, 5, 3 }, is_sorted_until ( beg, end, std::less<int>()) would return an iterator pointing at the second 3.

Given the sequence { 1, 2, 3, 4, 5, 9 }, is_sorted_until ( beg, end, std::less<int>()) would return end.

There are also a set of "wrapper functions" for is_ordered which make it easy to see if an entire sequence is ordered. These functions return a boolean indicating success or failure rather than an iterator to where the out of order items were found.

To test if a sequence is increasing (each element at least as large as the preceding one):

namespace boost { namespace algorithm {
	template <typename Iterator>
	bool is_increasing ( Iterator first, Iterator last );
	
	template <typename R>
	bool is_increasing ( const R &range );
}}

To test if a sequence is decreasing (each element no larger than the preceding one):

namespace boost { namespace algorithm {
	template <typename ForwardIterator>
	bool is_decreasing ( ForwardIterator first, ForwardIterator last );
	
	template <typename R>
	bool is_decreasing ( const R &range );
}}

To test if a sequence is strictly increasing (each element larger than the preceding one):

namespace boost { namespace algorithm {
	template <typename ForwardIterator>
	bool is_strictly_increasing ( ForwardIterator first, ForwardIterator last );
	
	template <typename R>
	bool is_strictly_increasing ( const R &range );
}}

To test if a sequence is strictly decreasing (each element smaller than the preceding one):

namespace boost { namespace algorithm {
	template <typename ForwardIterator>
	bool is_strictly_decreasing ( ForwardIterator first, ForwardIterator last );
	
	template <typename R>
	bool is_strictly_decreasing ( const R &range );
}}

Complexity: Each of these calls is just a thin wrapper over is_sorted, so they have the same complexity as is_sorted.

Notes

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