std::ranges::set_difference, std::ranges::set_difference_result

来自cppreference.com
< cpp‎ | algorithm‎ | ranges
 
 
算法库
受约束算法及范围上的算法 (C++20)
受约束算法: std::ranges::copy, std::ranges::sort, ...
执行策略 (C++17)
不修改序列的操作
(C++11)(C++11)(C++11)
(C++17)
修改序列的操作
Partitioning operations
划分操作
排序操作
(C++11)
二分搜索操作
集合操作(在已排序范围上)
堆操作
(C++11)
最小/最大操作
(C++11)
(C++17)

排列
数值运算
未初始化存储上的操作
(C++17)
(C++17)
(C++17)
C 库
 
受约束算法
不修改序列的操作
修改序列的操作
划分操作
排序操作
二分搜索操作
集合操作(在已排序范围上)
ranges::set_difference
堆操作
最小/最大操作
排列
未初始化存储上的操作
返回类型
 
在标头 <algorithm> 定义
调用签名
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 set_difference_result<I1, O>
          set_difference( I1 first1, S1 last1, I2 first2, S2 last2,
                          O result, Comp comp = {},

                          Proj1 proj1 = {}, Proj2 proj2 = {} );
(1) (C++20 起)
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 set_difference_result<ranges::borrowed_iterator_t<R1>, O>
          set_difference( R1&& r1, R2&& r2, O result, Comp comp = {},

                          Proj1 proj1 = {}, Proj2 proj2 = {} );
(2) (C++20 起)
辅助类型
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。

  • 输入范围未分别相对于 compproj1proj2 排序,或
  • 结果范围与任一输入范围重叠,

则行为未定义。

1) 用给定的二元比较函数 comp 比较元素。
2)(1) ,但以 r1 为第一范围,并以 r2 为第二范围,如同以 ranges::begin(r1)first1 ,以 ranges::end(r1)last1 ,以 ranges::begin(r2)first2 ,并以 ranges::end(r2)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,或者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
N
2
分别为 ranges::distance(first1, last1)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

计算两个集合的并集
(niebloid)
计算两个集合的交集
(niebloid)
计算两个集合的对称差
(niebloid)
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)