std::ranges::set_union, std::ranges::set_union_result - cppreference.com
若在 [first1, last1) 中找到某元素 m 次并在 [first2, last2) 中找到 n 次,则将从 [first1, last1) 复制 m 个元素到 result,保持顺序,然后将从 [first2, last2) 复制恰好 max(n-m, 0) 个元素到 result,亦保持顺序。
2) 同 (1),但以 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。
至多比较和应用每个投影 2·(N1+N2)-1 次,其中 N1 与 N2 分别为 ranges::distance(first1, last1) 与 ranges::distance(first2, last2)。
此算法进行与 ranges::merge 所做的类似的任务。两者都消费有序的输入范围并产生拥有来自两个范围的元素的有序输出。这两个算法间的区别在于如何处理来自两个输入范围的比较等价的值(见可小于比较 (LessThanComparable) 上的注解)。若任何等价的值在第一范围中出现 n 次,在第二范围出现 m 次,则 ranges::merge 将输出所有 n+m 次出现,而 ranges::set_union 则只输出 std::max(n, m) 次。故 ranges::merge 恰好输出 (N1+N2) 个值,而 ranges::set_union 可能产生较少的值。
struct set_union_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_union_result<I1, I2, O> operator()(I1 first1, S1 last1, I2 first2, S2 last2, O result, Comp comp = {}, Proj1 proj1 = {}, Proj2 proj2 = {}) const { for (; !(first1 == last1 or first2 == last2); ++result) { if (std::invoke(comp, std::invoke(proj1, *first1), std::invoke(proj2, *first2))) { *result = *first1; ++first1; } else if (std::invoke(comp, std::invoke(proj2, *first2), std::invoke(proj1, *first1))) { *result = *first2; ++first2; } else { *result = *first1; ++first1; ++first2; } } auto res1 = ranges::copy(std::move(first1), std::move(last1), std::move(result)); auto res2 = ranges::copy(std::move(first2), std::move(last2), std::move(res1.out)); return {std::move(res1.in), std::move(res2.in), std::move(res2.out)}; } 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_union_result<ranges::borrowed_iterator_t<R1>, ranges::borrowed_iterator_t<R2>, 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_union_fn set_union {};
#include <algorithm> #include <iostream> #include <iterator> #include <vector> void print(const auto& in1, const auto& in2, auto first, auto last) { std::cout << "{ "; for (const auto& e : in1) std::cout << e << ' '; std::cout << "} ∪ { "; for (const auto& e : in2) std::cout << e << ' '; std::cout << "} =\n{ "; while (!(first == last)) std::cout << *first++ << ' '; std::cout << "}\n\n"; } int main() { std::vector<int> in1, in2, out; in1 = {1, 2, 3, 4, 5}; in2 = { 3, 4, 5, 6, 7}; out.resize(in1.size() + in2.size()); const auto ret = std::ranges::set_union(in1, in2, out.begin()); print(in1, in2, out.begin(), ret.out); in1 = {1, 2, 3, 4, 5, 5, 5}; in2 = { 3, 4, 5, 6, 7}; out.clear(); out.reserve(in1.size() + in2.size()); std::ranges::set_union(in1, in2, std::back_inserter(out)); print(in1, in2, out.cbegin(), out.cend()); }
输出:
{ 1 2 3 4 5 } ∪ { 3 4 5 6 7 } =
{ 1 2 3 4 5 6 7 }
{ 1 2 3 4 5 5 5 } ∪ { 3 4 5 6 7 } =
{ 1 2 3 4 5 5 5 6 7 }