std::ranges::adjacent_find
来自cppreference.com
| 定义于头文件 <algorithm>
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| 调用签名 |
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| template< std::forward_iterator I, std::sentinel_for<I> S, class Proj = std::identity, std::indirect_binary_predicate< |
(1) | (C++20 起) |
| template< ranges::forward_range R, class Proj = std::identity, std::indirect_binary_predicate< |
(2) | (C++20 起) |
搜索范围 [first, last) 中二个连续的相等元素。
1) 用
pred （在以投影 proj 投影后）比较元素。2) 同 (1) ，但以
r 为源范围，如同以 ranges::begin(r) 为 first 并以 ranges::end(r) 为 last 。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，或以某些特殊编译器扩展实现。
参数
| first, last | - | 要检验的范围 |
| r | - | 要检验的范围 |
| pred | - | 应用到投影后元素的谓词 |
| proj | - | 应用到元素的投影 |
返回值
指向首对等同元素的前一者的迭代器，即首个使得 bool(std::invoke(pred, std::invoke(proj1, *it), std::invoke(proj, *(it + 1)))) 为 true 的 it 。
若找不到这种元素，则返回等于 last 的迭代器。
复杂度
准确应用 min((result-first)+1, (last-first)-1) 次谓词与投影，其中 result 是返回值。
可能的实现
struct adjacent_find_fn { template< std::forward_iterator I, std::sentinel_for<I> S, class Proj = std::identity, std::indirect_binary_predicate< std::projected<I, Proj>, std::projected<I, Proj>> Pred = ranges::equal_to > constexpr I operator()( I first, S last, Pred pred = {}, Proj proj = {} ) const { if (first == last) { return first; } auto next = ranges::next(first); for (; next != last; ++next, ++first) { if (std::invoke(pred, std::invoke(proj, *first), std::invoke(proj, *next))) { return first; } } return first; } template< ranges::forward_range R, class Proj = std::identity, std::indirect_binary_predicate< std::projected<ranges::iterator_t<R>, Proj>, std::projected<ranges::iterator_t<R>, Proj>> Pred = ranges::equal_to > constexpr ranges::borrowed_iterator_t<R> operator()( R&& r, Pred pred = {}, Proj proj = {} ) const { return (*this)(ranges::begin(r), ranges::end(r), std::ref(pred), std::ref(proj)); } }; inline constexpr adjacent_find_fn adjacent_find; |
示例
运行此代码
#include <algorithm> #include <iostream> #include <vector> #include <functional> int main() { std::vector<int> v1{0, 1, 2, 3, 40, 40, 41, 41, 5}; // ^^ ^^ namespace ranges = std::ranges; auto i1 = ranges::adjacent_find(v1.begin(), v1.end()); if (i1 == v1.end()) { std::cout << "No matching adjacent elements\n"; } else { std::cout << "The first adjacent pair of equal elements is at [" << ranges::distance(v1.begin(), i1) << "] == " << *i1 << '\n'; } auto i2 = ranges::adjacent_find(v1, ranges::greater()); if (i2 == v1.end()) { std::cout << "The entire vector is sorted in ascending order\n"; } else { std::cout << "The last element in the non-decreasing subsequence is at [" << ranges::distance(v1.begin(), i2) << "] == " << *i2 << '\n'; } }
输出：
The first adjacent pair of equal elements is at [4] == 40 The last element in the non-decreasing subsequence is at [7] == 41
参阅
| (C++20) |
移除范围中的连续重复元素 (niebloid) |
| 查找首对相邻的相同（或满足给定谓词的）元素 (函数模板) |