如何在c ++中找到字符串数组中字符串的最大出现次数

Question

我有一个类似的字符串数组

{"A", "B", "AA", "ABB", "B", "ABB", "B"}

如何在 c++ 中找到像这样的数组中出现次数最多的字符串（"B"）？

谢谢

Answer 1

如果数组元素的类型为std::string，那么在不使用额外内存资源的情况下，查找数组中出现频率更高的元素的函数可以如下所示，如下面的演示程序所示。

#include <iostream>
#include <string>

size_t max_occurence( const std::string a[], size_t n )
{
    size_t max = 0;
    size_t max_count = 0;
    
    for ( size_t i = 0; i != n; i++ )
    {
        size_t j = 0;
        
        while ( j != i && a[j] != a[i] ) ++j;
        
        if ( j == i )
        {
            size_t count = 1;
            while ( ++j != n )
            {
                if ( a[j] == a[i] ) ++count;
            }
            
            if ( max_count < count )
            {
                max = i;
                max_count = count;
            }
        }
    }
    
    return max;
}

int main() 
{
    std::string a[] = { "A", "B", "AA", "ABB", "B", "ABB", "B" };
    
    auto max = max_occurence( a, sizeof( a ) / sizeof( *a ) );
    
    std::cout << "The most often encountered string is " << a[max] << '\n';

    return 0;
}

程序输出是

The most often encountered string is B

另一种方法是编写通用模板函数。例如

#include <iostream>
#include <string>
#include <functional>
#include <iterator>

template <typename ForwardIterator, 
          typename BinaryPredicate = std::equal_to<typename std::iterator_traits<ForwardIterator>::value_type>>
ForwardIterator max_occurence( ForwardIterator first, 
                               ForwardIterator last, 
                               BinaryPredicate binary_predicate = BinaryPredicate() )
{
    auto max = first;
    typename std::iterator_traits<ForwardIterator>::difference_type max_count = 0;
    
    for ( auto current = first; current != last; ++current )
    {
        auto prev = first;
        
        while ( prev != current && !binary_predicate( *prev, *current ) ) ++prev;
        
        if ( prev == current )
        {
            typename std::iterator_traits<ForwardIterator>::difference_type count = 1;
            
            while ( ++prev != last )
            {
                if ( binary_predicate( *prev, *current ) ) ++count;
            }
            
            if ( max_count < count )
            {
                max = current;
                max_count = count;
            }
        }
    }
    
    return max;
}


int main() 
{
    std::string a[] = { "A", "B", "AA", "ABB", "B", "ABB", "B" };
    
    auto max = max_occurence( std::begin( a ), std::end( a ) );
    
    std::cout << "The most often encountered string is " << *max << '\n';

    return 0;
}

程序输出同上图

The most often encountered string is B

Answer 2

这是一个使用 std::map 的示例。

typedef std::map<std::string, int> Frequency_Map;
int main()
{
  const std::vector<std::string> test_data =
  {
      "A", "B", "AA", "ABB", "B", "ABB", "B",
  };

  Frequency_Map    frequency_table;
  std::string s;
  const size_t length = test_data.length();
  for (unsigned int i = 0; i < length; ++i)
  {
      Frequency_Map::iterator iter(frequency_table.find(test_data[i]);
      if (iter != frequency_table.end())
      {
          ++(iter->second);
      }
      else
      {
          frequency_table[test_data[i]] = 1;
      }
  }

  for (Frequency_Map::iterator iter = frequency_table.begin();
       iter != frequency_table.end();
       ++iter)
  {
      std::cout << iter->first << "    " << iter->second << "\n";
  }
  return 0;
}

上面的代码使用std::map 构建了一个频率表，然后输出字符串及其频率。上面的代码可以很容易地修改为找到具有最大（最大）频率的字符串。