std::set_symmetric_difference
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| Defined in header <algorithm>
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| template< class InputIt1, class InputIt2, class OutputIt > OutputIt set_symmetric_difference( InputIt1 first1, InputIt1 last1, |
(1) | |
| template< class InputIt1, class InputIt2, class OutputIt, class Compare > |
(2) | |
中未發現的排序範圍
[first1, last1)和元素中未發現的範圍內開始，[first2, last2)排序範圍[first2, last2)從排序後的範圍[first1, last1)從排序後的範圍d_first的元素複製。由此獲得的範圍也進行排序。預計兩個輸入範圍進行排序與operator<的第一個版本，第二個版本需要進行排序的比較函數comp。如果發現一些元件m次[first1, last1)和n[first2, last2)次，將被複制到d_first完全相同std::abs(m-n)倍。如果m>n，那麼最後這些元素被複制m-n[first1,last1)，否則最後n-m元素複製[first2,last2)。由此獲得的範圍不能重疊的輸入範圍.Original:
Copies the elements from the sorted range
[first1, last1) which are not found in the sorted range [first2, last2) and the elements from the sorted range [first2, last2) which are not found in the sorted range [first1, last1) to the range beginning at d_first. The resulting range is also sorted. 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), it will be copied to d_first exactly std::abs(m-n) times. If m>n, then the last m-n of those elements are copied from [first1,last1), otherwise the last n-m elements are copied from [first2,last2). The resulting range cannot overlap with either of the input ranges.The text has been machine-translated via Google Translate.
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目錄 |
[编辑] 參數
| first1, last1 | - | 第一次排序範圍內的元素
Original: the first sorted range of elements The text has been machine-translated via Google Translate. You can help to correct and verify the translation. Click here for instructions. | |||||||||
| first2, last2 | - | 所述第二排序條件範圍內的元素
Original: the second sorted range of elements 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
過去的結尾的Iterator構建範圍.
Original:
Iterator past the end of the constructed range.
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[编辑] 複雜性
最2·(N1+N2-1)比較，其中N1= std::distance(first1, last1)和N2= std::distance(first2, last2).
Original:
At most 2·(N1+N2-1) comparisons, where N1 = std::distance(first1, last1) and N2 = std::distance(first2, last2).
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[编辑] 可能的實現
| First version |
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template<class InputIt1, class InputIt2, class OutputIt> OutputIt set_difference(InputIt1 first1, InputIt1 last1, InputIt2 first2, InputIt2 last2, OutputIt d_first) { while (first1 != last1) { if (first2 == last2) return std::copy(first1, last1, d_first); if (*first1 < *first2) { *d_first++ = *first1++; } else { if (*first2 < *first1) { *d_first++ = *first2; } else { ++first1; } ++first2; } } return std::copy(first2, last2, d_first); } |
| Second version |
template<class InputIt1, class InputIt2, class OutputIt, class Compare> OutputIt set_difference(InputIt1 first1, InputIt1 last1, InputIt2 first2, InputIt2 last2, OutputIt d_first, Compare comp) { while (first1 != last1) { if (first2 == last2) return std::copy(first1, last1, d_first); if (comp(*first1, *first2)) { *d_first++ = *first1++; } else { if (comp(*first2, *first1)) { *d_first++ = *first2; } else { ++first1; } ++first2; } } return std::copy(first2, last2, d_first); } |
[编辑] 為例
#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_symmetric_difference( v1.begin(), v1.end(), v2.begin(), v2.end(), std::back_inserter(v_intersection)); for(int n : v_intersection) std::cout << n << ' '; }
Output:
1 2 3 4 6 8 9 10
[编辑] 另請參閱
| 返回true，如果一組的一個子集的另一個 Original: returns true if one set is a subset of another The text has been machine-translated via Google Translate. You can help to correct and verify the translation. Click here for instructions. （函數模板） | |
| 計算兩個集合之間的差異 Original: computes the difference between two sets The text has been machine-translated via Google Translate. You can help to correct and verify the translation. Click here for instructions. （函數模板） | |
| computes the union of two sets （函數模板） | |
| 計算兩個集合的交集 Original: computes the intersection of two sets The text has been machine-translated via Google Translate. You can help to correct and verify the translation. Click here for instructions. （函數模板） | |
