std::ranges::set_difference, std::ranges::set_difference_result
定义于头文件 <algorithm>
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调用签名 |
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template< std::input_iterator I1, std::sentinel_for<I1> S1, std::input_iterator I2, std::sentinel_for<I2> S2, |
(1) | (C++20 起) |
template< ranges::input_range R1, ranges::input_range R2, std::weakly_incrementable O, class Comp = ranges::less, |
(2) | (C++20 起) |
辅助类型 |
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template<class I, class O> using set_difference_result = ranges::in_out_result<I, O>; |
(3) | (C++20 起) |
从已排序的输入范围 [first1, last1)
赋值在已排序的输入范围 [first2, last2)
中找不到的元素到始于 result
的输出范围。
若
- 输入范围未分别相对于
comp
与proj1
或proj2
排序,或 - 结果范围与任一输入范围重叠,
则行为未定义。
comp
比较元素。r1
为第一范围,并以 r2
为第二范围,如同以 ranges::begin(r1) 为 first1
,以 ranges::end(r1) 为 last1
,以 ranges::begin(r2) 为 first2
,并以 ranges::end(r2) 为 last2
。此页面上描述的仿函数实体是 niebloid ,即:
实际上,它们能以函数对象,或以某些特殊编译器扩展实现。
参数
first1, last1 | - | 第一已排序输入范围 |
first2, last2 | - | 第二已排序输入范围 |
r1 | - | 第一已排序输入范围 |
r2 | - | 第二已排序输入范围 |
result | - | 目标范围的起始 |
comp | - | 应用到投影后元素的比较器 |
proj1 | - | 应用到第一范围元素的投影 |
proj2 | - | 应用到第二范围元素的投影 |
返回值
{last1, result_last} ,其中 result_last 为被构造范围末尾。
复杂度
至多比较和应用每个投影 2·(N
1+N
2)-1 次,其中 N
1 与 N
2 分别为 ranges::distance(first1, last1) 与 ranges::distance(first2, last2) 。
可能的实现
struct set_difference_fn { template< std::input_iterator I1, std::sentinel_for<I1> S1, std::input_iterator I2, std::sentinel_for<I2> S2, std::weakly_incrementable O, class Comp = ranges::less, class Proj1 = std::identity, class Proj2 = std::identity > requires std::mergeable<I1, I2, O, Comp, Proj1, Proj2> constexpr ranges::set_difference_result<I1, O> operator()( I1 first1, S1 last1, I2 first2, S2 last2, O result, Comp comp = {}, Proj1 proj1 = {}, Proj2 proj2 = {} ) const { while (!(first1 == last1 or first2 == last2)) { if (std::invoke(comp, std::invoke(proj1, *first1), std::invoke(proj2, *first2))) { *result = *first1; ++first1; ++result; } else if (std::invoke(comp, std::invoke(proj2, *first2), std::invoke(proj1, *first1))) { ++first2; } else { ++first1; ++first2; } } return ranges::copy(std::move(first1), std::move(last1), std::move(result)); } template< ranges::input_range R1, ranges::input_range R2, std::weakly_incrementable O, class Comp = ranges::less, class Proj1 = std::identity, class Proj2 = std::identity > requires std::mergeable<ranges::iterator_t<R1>, ranges::iterator_t<R2>, O, Comp, Proj1, Proj2> constexpr ranges::set_difference_result< ranges::borrowed_iterator_t<R1>, O> operator()( R1&& r1, R2&& r2, O result, Comp comp = {}, Proj1 proj1 = {}, Proj2 proj2 = {} ) const { return (*this)(ranges::begin(r1), ranges::end(r1), ranges::begin(r2), ranges::end(r2), std::move(result), std::move(comp), std::move(proj1), std::move(proj2)); } }; inline constexpr set_difference_fn set_difference{}; |
示例
#include <algorithm> #include <cassert> #include <iostream> #include <iterator> #include <string_view> #include <vector> auto print = [](const auto& v, std::string_view end = "") { for (std::cout << "{ "; auto i : v) std::cout << i << ' '; std::cout << "} " << end; }; struct Order // 拥有非常有趣的数据的类 { int order_id; friend std::ostream& operator<<(std::ostream& os, const Order& ord) { return os << "{" << ord.order_id << "},"; } }; int main() { const auto v1 = {1, 2, 5, 5, 5, 9}; const auto v2 = {2, 5, 7}; std::vector<int> diff; std::ranges::set_difference(v1, v2, std::back_inserter(diff)); print(v1, "∖ "); print(v2, "= "); print(diff, "\n"); // 我们想知道哪些顺序在旧和新状态之间“改变了”: const std::vector<Order> old_orders { {1}, {2}, {5}, {9}, }; const std::vector<Order> new_orders { {2}, {5}, {7}, }; std::vector<Order> cut_orders(old_orders.size() + new_orders.size()); auto [old_orders_end, cut_orders_last] = std::ranges::set_difference(old_orders, new_orders, cut_orders.begin(), {}, &Order::order_id, &Order::order_id); assert(old_orders_end == old_orders.end()); std::cout << "old orders = "; print(old_orders, "\n"); std::cout << "new orders = "; print(new_orders, "\n"); std::cout << "cut orders = "; print(cut_orders, "\n"); cut_orders.erase(cut_orders_last, end(cut_orders)); std::cout << "cut orders = "; print(cut_orders, "\n"); }
输出:
{ 1 2 5 5 5 9 } ∖ { 2 5 7 } = { 1 5 5 9 } old orders = { {1}, {2}, {5}, {9}, } new orders = { {2}, {5}, {7}, } cut orders = { {1}, {9}, {0}, {0}, {0}, {0}, {0}, } cut orders = { {1}, {9}, }
参阅
(C++20) |
计算两个集合的并集 (niebloid) |
(C++20) |
计算两个集合的交集 (niebloid) |
计算两个集合的对称差 (niebloid) | |
(C++20) |
若一个序列是另一个的子列则返回 true (niebloid) |
计算两个集合的差集 (函数模板) |