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std::lexicographical_compare

Материал из cppreference.com
 
 
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Модифицирующие линейные операции
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Множества (на отсортированных промежутках)
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lexicographical_compare
is_permutation(C++11)
next_permutation
prev_permutation
Числовые операции
Библиотека C
 
Заголовочный файл <algorithm>
template< class InputIt1, class InputIt2 >

bool lexicographical_compare( InputIt1 first1, InputIt1 last1,

                              InputIt2 first2, InputIt2 last2 );
(1)
template< class InputIt1, class InputIt2, class Compare >

bool lexicographical_compare( InputIt1 first1, InputIt1 last1,
                              InputIt2 first2, InputIt2 last2,

                              Compare comp );
(2)
Проверки, если первый [first1, last1) диапазоне лексикографически меньше', чем второй [first2, last2) диапазоне. Первый вариант используется operator< для сравнения элементов, вторая версия использует данную функцию сравнения comp.
Original:
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.
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Лексикографические сравнения операции со следующими свойствами:
Original:
Lexicographical comparison is a operation with the following properties:
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  • Два диапазонов по сравнению элемент за элементом.
    Original:
    Two ranges are compared element by element.
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  • Первый элемент несовпадение определяет, какой диапазон является лексикографически меньше' или' больше, чем другие.
    Original:
    The first mismatching element defines which range is lexicographically less or greater than the other.
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  • Если диапазон является префиксом другой, более короткий диапазон лексикографически меньше', чем другие.
    Original:
    If one range is a prefix of another, the shorter range is lexicographically less than the other.
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  • Если два диапазона имеют эквивалентные элементы и той же длины, то диапазоны лексикографически' равны.
    Original:
    If two ranges have equivalent elements and are of the same length, then the ranges are lexicographically equal.
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  • Пустой диапазон лексикографически меньше', чем любая непустая диапазон.
    Original:
    An empty range is lexicographically less than any non-empty range.
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  • Два пустых диапазонах лексикографически' равны.
    Original:
    Two empty ranges are lexicographically equal.
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Содержание

[править] Параметры

first1, last1 -
первый ряд элементов для изучения
Original:
the first range of elements to examine
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first2, last2 -
второй ряд элементов для изучения
Original:
the second range of elements to examine
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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 InputIt1 and InputIt2 can be dereferenced and then implicitly converted to  Type1 and  Type2 respectively. ​

Type requirements
-
InputIt1, InputIt2 must meet the requirements of InputIterator.

[править] Возвращаемое значение

true если первый диапазон лексикографически меньше', чем вторая.
Original:
true if the first range is lexicographically less than the second.
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[править] Сложность

В большинстве 2·min(N1, N2) применения операции сравнения, где N1 = std::distance(first1, last1) и N2 = std::distance(first2, last2).
Original:
At most 2·min(N1, N2) applications of the comparison operation, where N1 = std::distance(first1, last1) and N2 = std::distance(first2, last2).
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[править] Возможная реализация

First version
template<class InputIt1, class InputIt2>
bool lexicographical_compare(InputIt1 first1, InputIt1 last1,
                             InputIt2 first2, InputIt2 last2)
{
    for ( ; (first1 != last1) && (first2 != last2); first1++, first2++ ) {
        if (*first1 < *first2) return true;
        if (*first2 < *first1) return false;
    }
    return (first1 == last1) && (first2 != last2);
}
Second version
template<class InputIt1, class InputIt2, class Compare>
bool lexicographical_compare(InputIt1 first1, InputIt1 last1,
                             InputIt2 first2, InputIt2 last2,
                             Compare comp)
{
    for ( ; (first1 != last1) && (first2 != last2); first1++, first2++ ) {
        if (comp(*first1, *first2)) return true;
        if (comp(*first2, *first1)) return false;
    }
    return (first1 == last1) && (first2 != last2);
}

[править] Пример

#include <algorithm>
#include <iostream>
#include <vector>
#include <cstdlib>
#include <ctime>
 
int main()
{
    std::vector<char> v1 {'a', 'b', 'c', 'd'};
    std::vector<char> v2 {'a', 'b', 'c', 'd'};
 
    std::srand(std::time(0));
    while (!std::lexicographical_compare(v1.begin(), v1.end(),
                                         v2.begin(), v2.end())) {
        for (auto c : v1) std::cout << c << ' ';
        std::cout << ">= ";
        for (auto c : v2) std::cout << c << ' ';
        std::cout << '\n';
 
        std::random_shuffle(v1.begin(), v1.end());
        std::random_shuffle(v2.begin(), v2.end());
    }
 
    for (auto c : v1) std::cout << c << ' ';
    std::cout << "< ";
    for (auto c : v2) std::cout << c << ' ';
    std::cout << '\n';
}

Возможный вывод:

a b c d >= a b c d
d a b c >= c b d a
b d a c >= a d c b
a c d b < c d a b