std::upper_bound
来自cppreference.com
该页由英文版wiki使用Google Translate机器翻译而来。
该翻译可能存在错误或用词不当。鼠标停留在文本上可以看到原版本。你可以帮助我们修正错误或改进翻译。参见说明请点击这里. |
在头文件 <algorithm> 中定义
|
||
template< class ForwardIt, class T > ForwardIt upper_bound( ForwardIt first, ForwardIt last, const T& value ); |
(1) | |
template< class ForwardIt, class T, class Compare > ForwardIt upper_bound( ForwardIt first, ForwardIt last, const T& value, Compare comp ); |
(2) | |
返回一个迭代器指向的第一个元素的范围内
[first, last)
是“更大的”比value
。 operator<的第一个版本使用比较的元素,第二个版本使用给定的比较函数comp
.原文:
Returns an iterator pointing to the first element in the range
[first, last)
that is greater than value
. The first version uses operator< to compare the elements, the second version uses the given comparison function comp
.目录 |
[编辑] 参数
first, last | - | 检查的元素
|
value | - | 值进行比较的元素
|
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: bool cmp(const Type1 &a, const Type2 &b); The signature does not need to have const &, but the function must not modify the objects passed to it. |
类型要求 | ||
-ForwardIt 必须满足 ForwardIterator 的要求。
|
[编辑] 返回值
迭代器指向的第一个元素是“更大的”比
value
,或last
如果没有找到这样的元素,.原文:
iterator pointing to the first element that is greater than
value
, or last
if no such element is found.[编辑] 可能的实现
版本一 |
---|
template<class ForwardIt, class T> ForwardIt upper_bound(ForwardIt first, ForwardIt last, const T& value) { ForwardIt it; std::iterator_traits<ForwardIt>::distance_type count, step; count = std::distance(first,last); while (count > 0) { it = first; step = count / 2; std::advance(it, step); if (!(value < *it)) { first = ++it; count -= step + 1; } else count = step; } return first; } |
版本二 |
template<class ForwardIt, class T, class Compare> ForwardIt upper_bound(ForwardIt first, ForwardIt last, const T& value, Compare comp) { ForwardIt it; std::iterator_traits<ForwardIt>::distance_type count, step; count = std::distance(first,last); while (count > 0) { it = first; step = count / 2; std::advance(it, step); if (!comp(value, *it)), first = ++it; count -= step + 1 } else count = step; } return first; } |
[编辑] 示例
#include <algorithm> #include <iostream> #include <iterator> #include <vector> int main() { std::vector<int> data = { 1, 1, 2, 3, 3, 3, 3, 4, 4, 4, 5, 5, 6 }; auto lower = std::lower_bound(data.begin(), data.end(), 4); auto upper = std::upper_bound(data.begin(), data.end(), 4); std::copy(lower, upper, std::ostream_iterator<int>(std::cout, " ")); }
输出:
4 4 4
[编辑] 复杂度
的对数
first
和last
之间的距离原文:
logarithmic in the distance between
first
and last
[编辑] 另请参阅
返回匹配特定键值的元素区间 (函数模板) | |
返回指向第一个不小于给定值的元素的迭代器 (函数模板) |