std::ranges::search_n
在标头 <algorithm> 定义
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调用签名 |
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(1) | ||
template< std::forward_iterator I, std::sentinel_for<I> S, class T, class Pred = ranges::equal_to, class Proj = std::identity > |
(C++20 起) (C++26 前) |
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template< std::forward_iterator I, std::sentinel_for<I> S, class Pred = ranges::equal_to, class Proj = std::identity, |
(C++26 起) | |
(2) | ||
template< ranges::forward_range R, class T, class Pred = ranges::equal_to, class Proj = std::identity > |
(C++20 起) (C++26 前) |
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template< ranges::forward_range R, class Pred = ranges::equal_to, class Proj = std::identity, |
(C++26 起) | |
[
first,
last)
中搜索首个 count 个元素的连续序列,其投影后的值按照二元谓词 pred 均等于给定的值 value。此页面上描述的函数式实体是 niebloid,即:
实践中,可以作为函数对象,或者用某些特殊编译器扩展实现它们。
参数
first, last | - | 要检验的元素范围(又称草堆) |
r | - | 要检验的元素范围(又称草堆) |
count | - | 要搜索的序列长度 |
value | - | 要搜索的值(又称针) |
pred | - | 比较投影后元素与 value 的二元谓词
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proj | - | 应用到要检验的范围的元素的投影 |
返回值
[
first,
last)
中指代找到的序列的一对迭代器的 std::ranges::subrange。
若找不到这种序列,则返回 std::ranges::subrange{last, last}。
若 count ≤ 0 则返回 std::ranges::subrange{first, first}。复杂度
线性:至多应用 ranges::distance(first, last) 次谓词与投影。
注解
若迭代器实现了 std::random_access_iterator 则实现可以提升平均搜索效率。
功能特性测试宏 | 值 | 标准 | 功能特性 |
---|---|---|---|
__cpp_lib_algorithm_default_value_type |
202403 | (C++26) | 算法中的列表初始化 |
可能的实现
struct search_n_fn { template<std::forward_iterator I, std::sentinel_for<I> S, class Pred = ranges::equal_to, class Proj = std::identity, class T = std::projected_value_t<I, Proj>> requires std::indirectly_comparable<I, const T*, Pred, Proj> constexpr ranges::subrange<I> operator()(I first, S last, std::iter_difference_t<I> count, const T& value, Pred pred = {}, Proj proj = {}) const { if (count <= 0) return {first, first}; for (; first != last; ++first) if (std::invoke(pred, std::invoke(proj, *first), value)) { I start = first; std::iter_difference_t<I> n{1}; for (;;) { if (n++ == count) return {start, std::next(first)}; // 找到 if (++first == last) return {first, first}; // 找不到 if (!std::invoke(pred, std::invoke(proj, *first), value)) break; // 不等于 value } } return {first, first}; } template<ranges::forward_range R, class Pred = ranges::equal_to, class Proj = std::identity, class T = std::projected_value_t<ranges::iterator_t<R>, Proj>> requires std::indirectly_comparable<ranges::iterator_t<R>, const T*, Pred, Proj> constexpr ranges::borrowed_subrange_t<R> operator()(R&& r, ranges::range_difference_t<R> count, const T& value, Pred pred = {}, Proj proj = {}) const { return (*this)(ranges::begin(r), ranges::end(r), std::move(count), value, std::move(pred), std::move(proj)); } }; inline constexpr search_n_fn search_n {}; |
示例
#include <algorithm> #include <cassert> #include <complex> #include <iomanip> #include <iostream> #include <iterator> #include <string> #include <vector> int main() { namespace ranges = std::ranges; static constexpr auto nums = {1, 2, 2, 3, 4, 1, 2, 2, 2, 1}; constexpr int count{3}; constexpr int value{2}; typedef int count_t, value_t; constexpr auto result1 = ranges::search_n ( nums.begin(), nums.end(), count, value ); static_assert // 找到 ( result1.size() == count && std::distance(nums.begin(), result1.begin()) == 6 && std::distance(nums.begin(), result1.end()) == 9 ); constexpr auto result2 = ranges::search_n(nums, count, value); static_assert // 找到 ( result2.size() == count && std::distance(nums.begin(), result2.begin()) == 6 && std::distance(nums.begin(), result2.end()) == 9 ); constexpr auto result3 = ranges::search_n(nums, count, value_t{5}); static_assert // 找不到 ( result3.size() == 0 && result3.begin() == result3.end() && result3.end() == nums.end() ); constexpr auto result4 = ranges::search_n(nums, count_t{0}, value_t{1}); static_assert // 找不到 ( result4.size() == 0 && result4.begin() == result4.end() && result4.end() == nums.begin() ); constexpr char symbol{'B'}; auto to_ascii = [](const int z) -> char { return 'A' + z - 1; }; auto is_equ = [](const char x, const char y) { return x == y; }; std::cout << "找到了子序列 " << std::string(count, symbol) << " 于 "; std::ranges::transform(nums, std::ostream_iterator<char>(std::cout, ""), to_ascii); std::cout << '\n'; auto result5 = ranges::search_n(nums, count, symbol, is_equ, to_ascii); if (not result5.empty()) std::cout << "找到其位置在 " << ranges::distance(nums.begin(), result5.begin()) << '\n'; std::vector<std::complex<double>> nums2{{4, 2}, {4, 2}, {1, 3}}; #ifdef __cpp_lib_algorithm_default_value_type auto it = ranges::search_n(nums2, 2, {4, 2}); #else auto it = ranges::search_n(nums2, 2, std::complex<double>{4, 2}); #endif assert(it.size() == 2); }
输出:
找到了子序列 BBB 于 ABBCDABBBA 找到其位置在 6
参阅
(C++20) |
查找首对相邻的相同(或满足给定谓词的)元素 (niebloid) |
(C++20)(C++20)(C++20) |
查找满足特定条件的的第一个元素 (niebloid) |
(C++20) |
查找特定范围中最后出现的元素序列 (niebloid) |
(C++20) |
查找范围中的任一元素 (niebloid) |
(C++20) |
若一个序列是另一个的子列则返回 true (niebloid) |
(C++20) |
寻找两个范围出现不同的首个位置 (niebloid) |
(C++20) |
搜索一个元素范围的首次出现 (niebloid) |
在范围中搜索首次出现的某元素一定量连续副本 (函数模板) |