计算字符串中某个字符串出现的次数

Question

计算字符串中子字符串的所有出现次数的最佳方法是什么？

示例：计算

Foo

 中 

FooBarFooBarFoo

的出现次数

Answer 1

一种方法是使用 std::string find 函数：

#include <string>
#include <iostream>
int main()
{
   int occurrences = 0;
   std::string::size_type pos = 0;
   std::string s = "FooBarFooBarFoo";
   std::string target = "Foo";
   while ((pos = s.find(target, pos )) != std::string::npos) {
          ++ occurrences;
          pos += target.length();
   }
   std::cout << occurrences << std::endl;

}

Answer 2

#include <iostream>
#include <string>

// returns count of non-overlapping occurrences of 'sub' in 'str'
int countSubstring(const std::string& str, const std::string& sub)
{
    if (sub.length() == 0) return 0;
    int count = 0;
    for (size_t offset = str.find(sub); offset != std::string::npos;
     offset = str.find(sub, offset + sub.length()))
    {
        ++count;
    }
    return count;
}

int main()
{
    std::cout << countSubstring("FooBarFooBarFoo", "Foo")    << '\n';

    return 0;
}

Answer 3

您应该为此使用KMP算法。 它在 O(M+N) 时间内解决了这个问题，其中 M 和 N 是两个字符串的长度。 欲了解更多信息- https://www.geeksforgeeks.org/Frequency-substring-string/

KMP 算法的作用是搜索字符串模式。当一个模式的子模式在子模式中出现多个时，它会使用该属性来提高时间复杂度，即使在最坏的情况下也是如此。

KMP 的时间复杂度为 O(n)。 查看详细算法： https://www.geeksforgeeks.org/kmp-algorithm-for-pattern-searching/

Answer 4

#include <iostream>
#include<string>
using namespace std;
int frequency_Substr(string str,string substr)
{
    int count=0;
    for (int i = 0; i <str.size()-1; i++)
    {
        int m = 0;
        int n = i;
        for (int j = 0; j < substr.size(); j++)
        {
            if (str[n] == substr[j])
            {
                m++;
            }
            n++;
        }
        if (m == substr.size())
        {
            count++;
        }
    
    }
    cout << "total number of time substring occur in string is " << count << endl;
    return count;
}
int main()
{
    string x, y;
    cout << "enter string" << endl;
    cin >> x;
    cout << "enter substring" << endl;
    cin >> y;
    frequency_Substr(x, y);
    return 0;
}

Answer 5

#include<iostream>
#include<string>
using namespace std;
int main()
{
    string s1,s2;
    int i=0;
    cout<<"enter the string"<<endl;
    getline(cin,s1);
    cout<<"enter the substring"<<endl;
    cin>>s2;
    int count=0;
    string::iterator it=s1.begin();
    while(it!=s1.end())
    {
        if(*it==s2[0])
        {
            int x =s1.find(s2);
            string subs=s1.substr(x,s2.size());
            if(s2==subs)
                count++;
        }
        ++it;
    
    }
    cout<<count<<endl;
    return 0;
}

Answer 6

在我的软件中，我正在使用此函数（使用嵌套迭代器）：

int Global::count (const std::string& str, const std::string& substr) {
  int nocc (0);
  if (substr.begin () == substr.end ()) {
    // error ("count : empty substr");
  }
  if (substr.size () <= str.size ()) {
    const auto ssiz (substr.size ());
    auto i0 (substr.begin ()), j0 (substr.end ()), k0 (i0);
    auto i (str.begin ()), k (i);
    auto end (str.end ());
    end -= ssiz;
    auto c0 (*substr.begin ());
    std::for_each (str.begin (), str.end (), [&c0, &nocc, &i, &i0, &j0, &k0, &k] (const auto& c) {
      if (c == c0) {
        k0 = i0; k = i;
        for (; k0 != j0; ++k0, ++k) {
          if (*k0 != *k) break;
        }
        if (k0 == j0) ++nocc;
      }
      ++i;
    });
  }
  return nocc;
}

用途：

#include <string>
#include "Global.hpp"

int main (int argc, char* argv []) {
  int nocc (0);
  std::string str ("FooBarFooBarFoo"), substr ("Foo");
  nocc = Global::count (str, substr);
  std::cout << "nocc == " << nocc << std::endl;
}