std::lexicographical_compare
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<metanoindex/>
<tbody> </tbody>| Definido no cabeçalho <algorithm>
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template< class InputIt1, class InputIt2 > bool lexicographical_compare( InputIt1 first1, InputIt1 last1, InputIt2 first2, InputIt2 last2 ); |
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template< class InputIt1, class InputIt2, class Compare > bool lexicographical_compare( InputIt1 first1, InputIt1 last1, InputIt2 first2, InputIt2 last2, Compare comp ); |
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Verifica se o
[first1, last1) primeira gama é lexicographically menos do que o [first2, last2) segunda faixa. A primeira versão utiliza operator< para comparar os elementos, a segunda versão utiliza a função de comparação dada 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.The text has been machine-translated via Google Translate.
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Lexicographical comparação é uma operação com as seguintes propriedades:
Original:
Lexicographical comparison is a operation with the following properties:
The text has been machine-translated via Google Translate.
You can help to correct and verify the translation. Click here for instructions.
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- Duas faixas são comparados elemento por elemento.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. - O elemento descasamento primeiro define qual faixa é lexicographically menos ou' maior que o outro.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. - Se um intervalo é um prefixo de outro, mais curto intervalo é lexicographically menos do que o outro.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. - Se duas cadeias têm elementos equivalentes e são do mesmo comprimento, então, as faixas são lexicographically igual.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. - Uma faixa vazia é lexicographically menos do que qualquer intervalo não vazio.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. - Duas faixas vazias são lexicographically' igual.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.
Parâmetros
| first1, last1 | - | a primeira gama de elementos para examinar
Original: the first range of elements to examine The text has been machine-translated via Google Translate. You can help to correct and verify the translation. Click here for instructions. |
| first2, last2 | - | o segundo intervalo de elementos para examinar
Original: the second range of elements to examine The text has been machine-translated via Google Translate. You can help to correct and verify the translation. Click here for instructions. |
| 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:
The signature does not need to have |
| Type requirements | ||
-InputIt1, InputIt2 must meet the requirements of InputIterator.
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Valor de retorno
true se a primeira faixa é lexicographically' menos do que a segunda.Original:
true if the first range is lexicographically less than the second.The text has been machine-translated via Google Translate.
You can help to correct and verify the translation. Click here for instructions.
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Complexidade
Na maioria das aplicações de 2·min(N1, N2) de a operação de comparação, onde
N1 = std::distance(first1, last1) e 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).The text has been machine-translated via Google Translate.
You can help to correct and verify the translation. Click here for instructions.
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Possível implementação
| First version |
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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);
}
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| 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);
}
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Exemplo
#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';
}
Potencial saída:
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