Question

有没有办法在容器中找到比较每N个元素的max元素并返回索引。使用STL，BOOST或......另一个lib？

每个N，我的意思是使用std :: max_element，但是将for的增加从++改为first + = n;

// based on std::max_element

#ifndef NEWTON_ALGORITHM_SEARCH_MAX_INDEX_HPP
#define NEWTON_ALGORITHM_SEARCH_MAX_INDEX_HPP

#include <iterator>

#include <newton/functional.hpp>

namespace newton {

  // SAME THAT STD::MAX_ELEMENT
  template<class ForwardIt, class Compare>
  const ForwardIt max_index(ForwardIt first, ForwardIt last, Compare comp)
  {

    if ( newton::equal(first, last) ) // newton::equal is basically equivalent to std::equal_to
      return last;

    ForwardIt largest = first;
    while ( newton::not_equal(++first, last) )
      if (comp(*largest, *first))
        largest = first;

    return largest;

  }

  // possible names
  // max_index_some
  // max_index_every_n
  // max_index__n

  template<class ForwardIt, class Size, class Compare>
  const ForwardIt max_index_every_n(ForwardIt first, ForwardIt last, Size n, Compare comp)
  {

    if ( newton::equal(first, last) )
      return last;

    ForwardIt largest = first;
    Size blocks = std::distance(first, last) / n; // integer blocks

    if ( newton::greater_equal(blocks, 1) ) {

      // if there are exacly N elements, can't sum example
      // v.size() = 10, first start in 0, so "0 += 10" is "10", but last index is "9"
      // but if mod >= 1, then index is at least 10, so can sum

      if ( newton::greater_equal( std::distance(first, last) % n, 1) ) {
        for (size_t i = 1; newton::less_equal(i, blocks); ++i, first += n) {

          if (comp(*largest, *first))
            largest = first;
        }
      }
      else {

        for (size_t i = 1; newton::less(i, blocks); ++i, first += n) {

          if (comp(*largest, *first))
            largest = first;
        }

      }

    }

    return largest;

  }

  template<class ForwardIt>
  const ForwardIt max_index(ForwardIt first, ForwardIt last)
  {
    return max_index(first, last, newton::structure::less());
  }
} // newton

如果没有，那么您的解决方案是否尝试将其包含在下一个STL版本中。 remember

Answer 1

使用range-v3，它将是：

auto r = v | ranges::view::stride(n);
auto it = ranges::max_element(r);

Answer 2

仅使用Boost（＆＃34;范围V2＆＃34;可以这么说）：

<强> Live On Coliru

#include <boost/range/adaptors.hpp>
#include <boost/range/algorithm.hpp>
#include <boost/range/algorithm_ext.hpp>
#include <iostream>
#include <vector>

int main() {
    std::vector<int> v(100);
    boost::iota(v, 0);

    int n = 17;
    auto it = boost::max_element(v | boost::adaptors::strided(n));

    std::cout << "max: " << *it << "\n";
}

打印

max: 85

替代用法

您不必使用范围算法等等。您也可以根据需要选择：

<强> Live On Coliru

#include <boost/range/adaptor/strided.hpp>
#include <iostream>
#include <vector>

int main() {
    std::vector<int> v { 1,2,3,4,5,6,7,8,9,10 };

    auto v_ = boost::adaptors::stride(v, 7);
    auto it = std::max_element(v_.begin(), v_.end());

    std::cout << "max: " << *it << "\n";
}

打印

max: 8

每n个元素c ++ max_element

2 个答案:

替代用法