std::set_intersection
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| Defined in header <algorithm>
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| template< class InputIt1, class InputIt2, class OutputIt > OutputIt set_intersection( InputIt1 first1, InputIt1 last1, |
(1) | |
| template< class InputIt1, class InputIt2, class OutputIt, class Compare > |
(2) | |
d_first中发现的两个排序范围的元素组成的[first1, last1)[first2, last2)。预计两个输入范围进行排序与operator<的第一个版本，第二个版本需要进行排序的比较函数comp。如果一些元素被发现m次[first1, last1)n在[first2, last2)的时候，第一std::min(m, n)元素将被复制到目的范围从第一个范围。等同元件的顺序被保留。由此获得的范围不能重叠的输入范围.d_first consisting of elements that are found in both sorted ranges [first1, last1) and [first2, last2). The first version expects both input ranges to be sorted with operator<, the second version expects them to be sorted with the given comparison function comp. If some element is found m times in [first1, last1) and n times in [first2, last2), the first std::min(m, n) elements will be copied from the first range to the destination range. The order of equivalent elements is preserved. The resulting range cannot overlap with either of the input ranges.You can help to correct and verify the translation. Click here for instructions.
目录 |
[编辑] 参数
| first1, last1 | - | 检查的元素
Original: the first range of elements to examine The text has been machine-translated via Google Translate. You can help to correct and verify the translation. Click here for instructions. | |||||||||
| first2, last2 | - | 第二次检查的元素
Original: the second range of elements to examine The text has been machine-translated via Google Translate. You can help to correct and verify the translation. Click here for instructions. | |||||||||
| comp | - | comparison function which returns true if the first argument is less than the second. The signature of the comparison function should be equivalent to the following:
The signature does not need to have const &, but the function must not modify the objects passed to it. | |||||||||
| Type requirements | |||||||||||
-InputIt1 must meet the requirements of InputIterator.
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-InputIt2 must meet the requirements of InputIterator.
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-OutputIt must meet the requirements of OutputIterator.
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[编辑] 返回值
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[编辑] 复杂性
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[编辑] 可能的实现
| First version |
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template<class InputIt1, class InputIt2, class OutputIt> OutputIt set_intersection(InputIt1 first1, InputIt1 last1, InputIt2 first2, InputIt2 last2, OutputIt d_first) { while (first1 != last1 && first2 != last2) { if (*first1 < *first2) { ++first1; } else { if (!(*first2 < *first1)) { *d_first++ = *first1++; } ++first2; } } return d_first; } |
| Second version |
template<class InputIt1, class InputIt2, class OutputIt, class Compare> OutputIt set_intersection(InputIt1 first1, InputIt1 last1, InputIt2 first2, InputIt2 last2, OutputIt d_first, Compare comp) { while (first1 != last1 && first2 != last2) { if (comp(*first1, *first2)) { ++first1; } else { if (!comp(*first2, *first1)) { *d_first++ = *first1++; } ++first2; } } return d_first; } |
[编辑] 为例
#include <iostream> #include <vector> #include <algorithm> #include <iterator> int main() { std::vector<int> v1{1,2,3,4,5,6,7,8}; std::vector<int> v2{ 5, 7, 9,10}; std::sort(v1.begin(), v1.end()); std::sort(v2.begin(), v2.end()); std::vector<int> v_intersection; std::set_intersection(v1.begin(), v1.end(), v2.begin(), v2.end(), std::back_inserter(v_intersection)); for(int n : v_intersection) std::cout << n << ' '; }
Output:
5 7
[编辑] 另请参阅
| computes the union of two sets （函数模板） | |
