std::find_end
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| Defined in header <algorithm>
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| template< class ForwardIt1, class ForwardIt2 > ForwardIt1 find_end( ForwardIt1 first, ForwardIt1 last, |
(1) | |
| template< class ForwardIt1, class ForwardIt2, class BinaryPredicate > ForwardIt1 find_end( ForwardIt1 first, ForwardIt1 last, |
(2) | |
[s_first, s_last)[first, last)。的第一个版本使用operator==比较的元素,第二个版本使用给定的二元谓词p. [s_first, s_last) in the range [first, last). The first version uses operator== to compare the elements, the second version uses the given binary predicate p. You can help to correct and verify the translation. Click here for instructions.
目录 |
[编辑] 参数
| first, last | - | 检查的元素
Original: the 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. | |||||||||
| s_first, s_last | - | 的元素的范围搜索
Original: the range of elements to search for The text has been machine-translated via Google Translate. You can help to correct and verify the translation. Click here for instructions. | |||||||||
| p | - | binary predicate which returns true if the elements should be treated as equal. The signature of the predicate 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 | |||||||||||
-ForwardIt1 must meet the requirements of ForwardIterator.
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-ForwardIt2 must meet the requirements of ForwardIterator.
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[编辑] 返回值
[s_first, s_last)[first, last).[s_first, s_last) in range [first, last).You can help to correct and verify the translation. Click here for instructions.
last返回。 (至 C++11)last is returned. (至 C++11)You can help to correct and verify the translation. Click here for instructions.
[s_first, s_last)是空的,或者如果没有这样的序列被发现,last返回。 (C++11 起)[s_first, s_last) is empty or if no such subsequence is found, last is returned. (C++11 起)You can help to correct and verify the translation. Click here for instructions.
[编辑] 复杂性
S*(N-S+1)比较S = distance(s_first, s_last)N = distance(first, last).S*(N-S+1) comparisons where S = distance(s_first, s_last) and N = distance(first, last).You can help to correct and verify the translation. Click here for instructions.
[编辑] 可能的实现
| First version |
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template<class ForwardIt1, class ForwardIt2> ForwardIt1 find_end(ForwardIt1 first, ForwardIt1 last, ForwardIt2 s_first, ForwardIt2 s_last) { if (s_first == s_last) return last; ForwardIt1 result = last; while (1) { ForwardIt1 new_result = std::search(first, last, s_first, s_last); if (new_result == last) { return result; } else { result = new_result; first = result; ++first; } } return result; } |
| Second version |
template<class ForwardIt1, class ForwardIt2, class BinaryPredicate> ForwardIt1 find_end(ForwardIt1 first, ForwardIt1 last, ForwardIt2 s_first, ForwardIt2 s_last, BinaryPredicate p) { if (s_first == s_last) return last; ForwardIt1 result = last; while (1) { ForwardIt1 new_result = std::search(first, last, s_first, s_last, p); if (new_result == last) { return result; } else { result = new_result; first = result; ++first; } } return result; } |
[编辑] 为例
find_end()搜索两个不同的数字序列.
find_end() to search for two different sequences of numbers.
You can help to correct and verify the translation. Click here for instructions.
#include <algorithm> #include <iostream> #include <vector> int main() { std::vector<int> v{1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4}; std::vector<int>::iterator result; std::vector<int> t1{1, 2, 3}; result = std::find_end(v.begin(), v.end(), t1.begin(), t1.end()); if (result == v.end()) { std::cout << "subsequence not found\n"; } else { std::cout << "last subsequence is at: " << std::distance(v.begin(), result) << "\n"; } std::vector<int> t2{4, 5, 6}; result = std::find_end(v.begin(), v.end(), t2.begin(), t2.end()); if (result == v.end()) { std::cout << "subsequence not found\n"; } else { std::cout << "last subsequence is at: " << std::distance(v.begin(), result) << "\n"; } }
Output:
last subsequence is at: 8 subsequence not found
[编辑] 另请参阅
| 发现两个相同的(或一些其他的关系)件彼此相邻 Original: finds two identical (or some other relationship) items adjacent to each other The text has been machine-translated via Google Translate. You can help to correct and verify the translation. Click here for instructions. (函数模板) | |
| (C++11) |
找到的第一个元素,满足特定的标准 Original: finds the first element satisfying specific criteria The text has been machine-translated via Google Translate. You can help to correct and verify the translation. Click here for instructions. (函数模板) |
| searches for any one of a set of elements (函数模板) | |
| 检索数据的一些连续的范围中的一个元素的副本 Original: searches for a number consecutive copies of an element in a range The text has been machine-translated via Google Translate. You can help to correct and verify the translation. Click here for instructions. (函数模板) | |