std::ranges::lexicographical_compare
来自cppreference.com
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| 在标头 <algorithm> 定义
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template< std::input_iterator I1, std::sentinel_for<I1> S1, std::input_iterator I2, std::sentinel_for<I2> S2, class Proj1 = std::identity, class Proj2 = std::identity, std::indirect_strict_weak_order< std::projected<I1, Proj1>, std::projected<I2, Proj2>> Comp = ranges::less > constexpr bool lexicographical_compare( I1 first1, S1 last1, I2 first2, S2 last2, Comp comp = {}, Proj1 proj1 = {}, Proj2 proj2 = {} ); |
(1) | (C++20 起) |
template< ranges::input_range R1, ranges::input_range R2, class Proj1 = std::identity, class Proj2 = std::identity, std::indirect_strict_weak_order< std::projected<ranges::iterator_t<R1>, Proj1>, std::projected<ranges::iterator_t<R2>, Proj2>> Comp = ranges::less > constexpr bool lexicographical_compare( R1&& r1, R2&& r2, Comp comp = {}, Proj1 proj1 = {}, Proj2 proj2 = {} ); |
(2) | (C++20 起) |
检查第一范围 [first1, last1) 是否按字典序小于第二范围 [first2, last2)。
1) 用给定的二元比较函数
comp 比较元素。2) 同 (1)，但以
r 为源范围，如同以 ranges::begin(r) 为 first 并以 ranges::end(r) 为 last。字典序比较是拥有下列属性的操作：
- 按元素逐个比较两个范围。
- 首个不匹配元素定义范围按字典序小于或大于另一个。
- 若一个范围是另一个的前缀，则较短的范围按字典序小于另一个。
- 若两个范围拥有等价的元素与相同长度，则范围按字典序相等。
- 空范围按字典序小于任何非空范围。
- 两个空范围按字典序相等。
此页面上描述的函数式实体是算法函数对象（非正式地称为 niebloid），即：
参数
| first1, last1 | - | 要检验的第一元素范围的迭代器-哨位对 |
| r1 | - | 要检验的第一元素范围 |
| first2, last2 | - | 要检验的第二元素范围的迭代器-哨位对 |
| r2 | - | 要检验的第二元素范围 |
| comp | - | 应用到投影后元素的比较函数 |
| proj1 | - | 应用到第一元素范围的投影 |
| proj2 | - | 应用到第二元素范围的投影 |
返回值
若第一范围按字典序小于第二范围则为 true 。
复杂度
至多应用 2·min(N1, N2) 次比较与对应的投影，其中 N1 = ranges::distance(first1, last1) 而 N2 = ranges::distance(first2, last2)。
可能的实现
struct lexicographical_compare_fn
{
template<std::input_iterator I1, std::sentinel_for<I1> S1,
std::input_iterator I2, std::sentinel_for<I2> S2,
class Proj1 = std::identity, class Proj2 = std::identity,
std::indirect_strict_weak_order<
std::projected<I1, Proj1>,
std::projected<I2, Proj2>> Comp = ranges::less>
constexpr bool operator()(I1 first1, S1 last1, I2 first2, S2 last2,
Comp comp = {}, Proj1 proj1 = {}, Proj2 proj2 = {}) const
{
for (; (first1 != last1) && (first2 != last2); ++first1, (void) ++first2)
{
if (std::invoke(comp, std::invoke(proj1, *first1), std::invoke(proj2, *first2)))
return true;
if (std::invoke(comp, std::invoke(proj2, *first2), std::invoke(proj1, *first1)))
return false;
}
return (first1 == last1) && (first2 != last2);
}
template<ranges::input_range R1, ranges::input_range R2,
class Proj1 = std::identity, class Proj2 = std::identity,
std::indirect_strict_weak_order<
std::projected<ranges::iterator_t<R1>, Proj1>,
std::projected<ranges::iterator_t<R2>, Proj2>> Comp = ranges::less>
constexpr bool operator()(R1&& r1, R2&& r2, Comp comp = {},
Proj1 proj1 = {}, Proj2 proj2 = {}) const
{
return (*this)(ranges::begin(r1), ranges::end(r1),
ranges::begin(r2), ranges::end(r2),
std::ref(comp), std::ref(proj1), std::ref(proj2));
}
};
inline constexpr lexicographical_compare_fn lexicographical_compare;
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示例
运行此代码
#include <algorithm>
#include <iostream>
#include <iterator>
#include <random>
#include <vector>
int main()
{
std::vector<char> v1 {'a', 'b', 'c', 'd'};
std::vector<char> v2 {'a', 'b', 'c', 'd'};
namespace ranges = std::ranges;
auto os = std::ostream_iterator<char>(std::cout, " ");
std::mt19937 g {std::random_device {}()};
while (not ranges::lexicographical_compare(v1, v2))
{
ranges::copy(v1, os);
std::cout << ">= ";
ranges::copy(v2, os);
std::cout << '\n';
ranges::shuffle(v1, g);
ranges::shuffle(v2, g);
}
ranges::copy(v1, os);
std::cout << "< ";
ranges::copy(v2, os);
std::cout << '\n';
}
可能的输出：
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
参阅
(C++20) |
判断两组元素是否相同 (算法函数对象) |
当一个范围字典序小于另一个时返回 true (函数模板) |