std::lexicographical_compare

[]

This page has been machine-translated from the English version of the wiki using Google Translate. The translation may contain errors and awkward wording. Hover over text to see the original version. You can help to fix errors and improve the translation. For instructions click here. Click here for the English version of this page

<metanoindex/>

• Deux gammes sont compares lment par lment .

  Original:

  Two ranges are compared element by element.

  The text has been machine-translated via Google Translate.
  You can help to correct and verify the translation. Click here for instructions.
• L'lment non-concordance premire dfinit ce qui est lexicographiquement gamme moins ou' plus que l'autre .

  Original:

  The first mismatching element defines which range is lexicographically less or greater than the other.

  The text has been machine-translated via Google Translate.
  You can help to correct and verify the translation. Click here for instructions.
• Si une plage est un prfixe d'un autre, plus courte porte est lexicographiquement moins que l'autre .

  Original:

  If one range is a prefix of another, the shorter range is lexicographically less than the other.

  The text has been machine-translated via Google Translate.
  You can help to correct and verify the translation. Click here for instructions.
• Si deux gammes prsentent des lments quivalents et sont de la mme longueur, puis les intervalles sont gaux' lexicographiquement .

  Original:

  If two ranges have equivalent elements and are of the same length, then the ranges are lexicographically equal.

  The text has been machine-translated via Google Translate.
  You can help to correct and verify the translation. Click here for instructions.
• Une plage vide est lexicographiquement moins que tout autre domaine non vide .

  Original:

  An empty range is lexicographically less than any non-empty range.

  The text has been machine-translated via Google Translate.
  You can help to correct and verify the translation. Click here for instructions.
• Deux plages vides sont lexicographiquement' galit .

  Original:

  Two empty ranges are lexicographically equal.

  The text has been machine-translated via Google Translate.
  You can help to correct and verify the translation. Click here for instructions.

Paramtres

Retourne la valeur

Complexit

Mise en uvre possible

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);
}

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);
}

Exemple

#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';
}

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