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std::lexicographical_compare

From cppreference.com
 
 
 
Defined in header <algorithm>
template< class InputIterator1, class InputIterator2 >

bool lexicographical_compare( InputIterator1 first1, InputIterator1 last1,

                              InputIterator2 first2, InputIterator2 last2 );
(1)
template< class InputIterator1, class InputIterator2, class Compare >

bool lexicographical_compare( InputIterator1 first1, InputIterator1 last1,
                              InputIterator2 first2, InputIterator2 last2,

                              Compare comp );
(2)

Checks if the first range [first1, last1) is lexicographically less than the second range [first2, last2). The first version uses operator< to compare the elements, the second version uses the given comparison function comp.

Lexicographical comparison is a operation with the following properties:

  • Two ranges are compared element by element.
  • The first mismatching element defines which range is lexicographically less or greater than the other.
  • If one range is a prefix of another, the shorter range is lexicographically less than the other.
  • If two ranges have equivalent elements and are of the same length, then the ranges are lexicographically equal.
  • An empty range is lexicographically less than any non-empty range.
  • Two empty ranges are lexicographically equal.

Contents

[edit] Parameters

first1, last1 - the first range of elements to examine
first2, last2 - the second range of elements to examine
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:

bool cmp(const Type1 &a, const Type2 &b);

The signature does not need to have const &, but the function must not modify the objects passed to it.
The types  Type1 and  Type2 must be such that objects of types InputIterator1 and InputIterator2 can be dereferenced and then implicitly converted to  Type1 and  Type2 respectively. ​

[edit] Return value

true if the first range is lexicographically less than the second.

[edit] Complexity

At most 2·min(N1, N2) applications of the comparison operation, where N1 = std::distance(first1, last1) and N2 = std::distance(first2, last2).

[edit] Possible implementation

[edit] Example