std::ranges::search_n
定義於標頭檔案 <algorithm> |
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呼叫簽名 (Call signature) |
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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) (直到 C++26) |
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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) (直到 C++26) |
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template< ranges::forward_range R, class Pred = ranges::equal_to, class Proj = std::identity, |
(C++26 起) | |
[
first,
last)
中搜索第一個由 count 個元素組成的序列,這些元素的投影值根據二元謂詞 pred 都等於給定的 value。本頁描述的類函式實體是 演算法函式物件(非正式地稱為 niebloids),即
目錄 |
[編輯] 引數
first, last | - | 定義要檢查的元素範圍(又稱乾草堆)的迭代器-哨兵對 |
r | - | 要檢查的元素範圍(又稱乾草堆) |
count | - | 要搜尋的序列長度 |
value | - | 要搜尋的值(又稱“針”) |
pred | - | 比較投影元素與 value 的二元謂詞 |
proj | - | 應用於要檢查的範圍元素的投影 |
[編輯] 返回值
[
first,
last)
內的一對迭代器,這些迭代器指定找到的子序列。如果未找到此類子序列,返回 std::ranges::subrange{last, last}。
如果 count <= 0,返回 std::ranges::subrange{first, first}。[編輯] 複雜度
線性:最多應用謂詞和投影 ranges::distance(first, last) 次。
[編輯] 注意
如果迭代器建模 std::random_access_iterator,實現可以提高搜尋的平均效率。
特性測試宏 | 值 | 標準 | 特性 |
---|---|---|---|
__cpp_lib_algorithm_default_value_type |
202403 |
(C++26) | 演算法的列表初始化 |
[編輯] 可能的實現
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)}; // found if (++first == last) return {first, first}; // not found if (!std::invoke(pred, std::invoke(proj, *first), value)) break; // not equ to 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 {}; |
[編輯] 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 // found ( 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 // found ( 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 // not found ( 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 // not found ( 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 << "Find a sub-sequence " << std::string(count, symbol) << " in the "; 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 << "Found at position " << 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); }
輸出
Find a sub-sequence BBB in the ABBCDABBBA Found at position 6
[編輯] 另請參閱
(C++20) |
尋找第一對相等的(或滿足給定謂詞的)相鄰項 (演算法函式物件) |
(C++20)(C++20)(C++20) |
尋找第一個滿足特定條件的元素 (演算法函式物件) |
(C++20) |
在特定範圍中尋找最後一次出現的元素序列 (演算法函式物件) |
(C++20) |
搜尋一組元素中的任何一個 (演算法函式物件) |
(C++20) |
如果一個序列是另一個序列的子序列,則返回 true (演算法函式物件) |
(C++20) |
尋找兩個範圍開始不同的第一個位置 (演算法函式物件) |
(C++20) |
搜尋一個範圍的元素首次出現的位置 (演算法函式物件) |
在一個範圍內搜尋一個元素的連續 N 次副本首次出現的位置 (函式模板) |