Question

我有一个排序的字符串向量，我试图找到向量中每个元素的共存：

V = {“AAA”，“AAA”，“AAA”，“BCA”，...}

int main()
{
      vector<string> vec;
      //for every word in the vector
      for(size_t i = 0; i < vec.size();i++)
       {

             int counter = 0;
              //loop through the vector and count the coocurrence of this word
             for(size_t j = 0; j < vec.size();j++)
              {
                 if(vec[i] == vec[j]) counter +=1;
              }

              cout << vec[i] << "    "<<counter <<ed,l
         }
}

复杂性是O（n ^ 2）对吗？我花了这么多时间才能找到解决问题的方法？

谢谢，

这就是编辑：

int main()
{
      vector<string> vec;
      //for every word in the vector
      for(size_t i = 0; i < vec.size();i++)
       {

             int counter = 0;
              //loop through the vector and count the coocurrence of this word
             for(size_t j = i+1; j < vec.size()-1;j++)
              {
                 if(vec[i] == vec[j]) counter +=1;
              }

              cout << vec[i] << "    "<<counter <<ed,l
         }
}

Answer 1

未经测试。我假设向量包含至少一个元素。

counter = 1
for(size_t i = 1; i < vec.size(); i++)
  {
    if(vec[i] == vec[i-1]) counter +=1;
    else 
      {
         std::cout << vec[i-1] << ", " << counter << std::endl;
         counter = 1;
      }
  }
std::cout << vec[i-1] << ", " << counter << std::endl;

这显然是O（n）。与您的代码略有不同：每个单词只打印一次。

Answer 2

经过测试，O（n），即使矢量未被排序或者它是空的，也能正常工作：

#include <iostream>
#include <vector>
#include <unordered_map>

int main()
{
    std::vector<std::string> v = { "aaa", "abc", "aaa", "def", "aaa", "aaa", "abc", "ghi" };
    std::unordered_map<std::string, int> m;

    for (std::vector<std::string>::iterator it = v.begin(); it != v.end(); it++)
        m[*it]++;

    for (std::unordered_map<std::string, int>::iterator it = m.begin(); it != m.end(); it++)
        std::cout << it->first << " -> " << it->second << std::endl;

    return 0;
}

或者，为了便于阅读，使用基于范围的循环重写了相应的片段（感谢Frerich Raabe）：

for (const auto it: v)
    m[it]++;

for (const auto it: m)
    std::cout << it.first << " -> " << it.second << std::endl;

计数共现排序的矢量字符串c ++

2 个答案: