字符串数组中出现次数最高的元素 [英] Element with highest occurence in array of strings

问题描述

我需要在字符串数组中找到出现次数最多的元素.我不确定该怎么做，因为我对此没有太多经验.我不知道指针/哈希表.

I need to find the element with the highest occurrence in an array of strings. I am not sure what to do, since I don't have much experience with this. I don't know pointers / hashtables.

我已经为整数完成了此操作，但是我无法使其适用于字符串.

I've already done this for integers, but I can't make it work for strings.

我的版本:

#include <iostream>
using namespace std;
int main(int argc, char *argv[])
{
    int a[]={1,2,3,4,4,4,5};
    int n = sizeof(a)/sizeof(int );
    int *b=new int [n];
    fill_n(b,n,0); // Put n times 0 in b

    int val=0; // Value that is the most frequent
    for (int i=0;i<n;i++)
        if( ++b[a[i]] >= b[val])
            val = a[i];

    cout<<val<<endl;
    delete[] b;
    return 0;
    }

对于在字符串数组中查找最频繁出现的元素的任何帮助，我们深表感谢！

Any help for finding the most frequently occurring element in the array of strings is appreciated!

推荐答案

您可以考虑使用向量，并使用地图进行计数:

You could consider using a vector of strings and to use a map to count occurrences:

#include <iostream>
#include <vector>
#include <map>

using namespace std;

int main(int argc, char *argv[])
{
  vector<string> a;
  map<string, int> m;

  // fill a with strings
  a.push_back("a");
  a.push_back("b");
  a.push_back("b");
  a.push_back("c");
  a.push_back("c");
  a.push_back("c");
  a.push_back("d");
  a.push_back("e");

  // count occurrences of every string
  for (int i = 0; i < a.size(); i++)
    {
      map<string, int>::iterator it = m.find(a[i]);

      if (it == m.end())
        m.insert(pair<string, int>(a[i], 1));

      else
        m[a[i]] += 1;
     }

  // find the max
  map<string, int>::iterator it = m.begin();
  for (map<string, int>::iterator it2 = m.begin(); it2 != m.end(); ++it2)
    {
      if (it2 -> second > it -> second)
      it = it2;
    }

  cout << it -> first << endl;  
  return 0;
} 

就优雅和效率而言，此解决方案可能不是最佳解决方案，但应该给出想法.

This solution is probably not the best one in terms of elegance and efficiency, however it should give the idea.

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