std::ranges::partition
来自 cppreference.com
在头文件 <algorithm> 中定义 |
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调用签名 |
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template< std::permutable I, std::sentinel_for<I> S, class Proj = std::identity, std::indirect_unary_predicate<std::projected<I, Proj>> Pred > |
(1) | (自 C++20 起) |
template< ranges::forward_range R, class Proj = std::identity, std::indirect_unary_predicate< |
(2) | (自 C++20 起) |
1) 以这样的方式重新排序范围
[
first,
last)
中的元素,使谓词 pred 返回 true 的所有元素的投影 proj 位于谓词 pred 返回 false 的元素的投影 proj 之前。元素的相对顺序不会被保留。此页面上描述的类似函数的实体是 *niebloids*,即
在实践中,它们可以实现为函数对象,或使用特殊的编译器扩展。
内容 |
[edit] 参数
first, last | - | 要重新排序的元素范围 |
r | - | 要重新排序的元素范围 |
pred | - | 要应用于投影元素的谓词 |
proj | - | 要应用于元素的投影 |
[edit] 返回值
一个子范围,从指向第二组第一个元素的迭代器开始,并以等于 last 的迭代器结束。 (2) 返回 std::ranges::dangling 如果 r 是非 borrowed_range
类型的右值。
[edit] 复杂度
给定 N = ranges::distance(first, last),谓词和投影正好应用 N 次。如果 I
模仿 ranges::bidirectional_iterator,最多交换 N / 2 次,否则最多交换 N 次。
[edit] 可能的实现
struct partition_fn { template<std::permutable I, std::sentinel_for<I> S, class Proj = std::identity, std::indirect_unary_predicate<std::projected<I, Proj>> Pred> constexpr ranges::subrange<I> operator()(I first, S last, Pred pred, Proj proj = {}) const { first = ranges::find_if_not(first, last, std::ref(pred), std::ref(proj)); if (first == last) return {first, first}; for (auto i = ranges::next(first); i != last; ++i) { if (std::invoke(pred, std::invoke(proj, *i))) { ranges::iter_swap(i, first); ++first; } } return {std::move(first), std::move(last)}; } template<ranges::forward_range R, class Proj = std::identity, std::indirect_unary_predicate< std::projected<ranges::iterator_t<R>, Proj>> Pred> requires std::permutable<ranges::iterator_t<R>> constexpr ranges::borrowed_subrange_t<R> operator()(R&& r, Pred pred, Proj proj = {}) const { return (*this)(ranges::begin(r), ranges::end(r), std::ref(pred), std::ref(proj)); } }; inline constexpr partition_fn partition; |
[edit] 示例
运行这段代码
#include <algorithm> #include <forward_list> #include <functional> #include <iostream> #include <iterator> #include <ranges> #include <vector> namespace ranges = std::ranges; template<class I, std::sentinel_for<I> S, class Cmp = ranges::less> requires std::sortable<I, Cmp> void quicksort(I first, S last, Cmp cmp = Cmp {}) { using reference = std::iter_reference_t<I>; if (first == last) return; auto size = ranges::distance(first, last); auto pivot = ranges::next(first, size - 1); ranges::iter_swap(pivot, ranges::next(first, size / 2)); auto tail = ranges::partition(first, pivot, [=](reference em) { return std::invoke(cmp, em, *pivot); // em < pivot }); ranges::iter_swap(pivot, tail.begin()); quicksort(first, tail.begin(), std::ref(cmp)); quicksort(ranges::next(tail.begin()), last, std::ref(cmp)); } int main() { std::ostream_iterator<int> cout {std::cout, " "}; std::vector<int> v {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}; std::cout << "Original vector: \t"; ranges::copy(v, cout); auto tail = ranges::partition(v, [](int i) { return i % 2 == 0; }); std::cout << "\nPartitioned vector: \t"; ranges::copy(ranges::begin(v), ranges::begin(tail), cout); std::cout << "│ "; ranges::copy(tail, cout); std::forward_list<int> fl {1, 30, -4, 3, 5, -4, 1, 6, -8, 2, -5, 64, 1, 92}; std::cout << "\nUnsorted list: \t\t"; ranges::copy(fl, cout); quicksort(ranges::begin(fl), ranges::end(fl), ranges::greater {}); std::cout << "\nQuick-sorted list: \t"; ranges::copy(fl, cout); std::cout << '\n'; }
可能的输出
Original vector: 0 1 2 3 4 5 6 7 8 9 Partitioned vector: 0 8 2 6 4 │ 5 3 7 1 9 Unsorted list: 1 30 -4 3 5 -4 1 6 -8 2 -5 64 1 92 Quick-sorted list: 92 64 30 6 5 3 2 1 1 1 -4 -4 -5 -8
[edit] 另请参阅
(C++20) |
复制一个范围,将元素分成两组 (niebloid) |
(C++20) |
确定范围是否由给定谓词划分 (niebloid) |
(C++20) |
将元素分成两组,同时保留其相对顺序 (niebloid) |
将元素范围分成两组 (函数模板) |