定位子向量< string>在另一个向量< string> [英] locate sub-vector<string> in another vector<string>

问题描述

我有一个向量< string> v1 = {A，B，C} 。我想检查在中是否包含 vector< string> v2 = {X，Y，A，B，C，D} 。

I have a vector<string> v1 = {"A","B","C"}. I want to check if v1 is included in vector<string> v2 = {"X","Y","A","B","C","D"}.

• 我可以使用 STL 吗？


• 我可以找到一个集合是否是另一个集合的子集不排序


• 如果发现子集只是一次，算法停止 if（counter == v1.size（））{ break;} 。您认为我应该允许它继续搜索，以防子集重复两次？

• Can I find if a set is a subset of the other by using STL?
• The vectors should not be sorted
• If the subset is found just once the algorithm stops " if(counter == v1.size()){break;}". Do you think that I should allow it to continue the search in case the subset is repeated twice?

#include <vector>
#include <iostream>
#include <algorithm>

using namespace std;

float wordOrder ( std::vector<string> v1, std::vector<string> v2 )
{
  //declare a vector that will be used as an index. If we find the element of v1 in v2 we insert 1
  std::vector<int> index ( v2.size(),0 );
  int counter = 0;
  int s = v1.size();
  //check if size of v1 less than size of V2
  if ( v1.size() <= v2.size() ) {

    for ( int i = 0; i < v1.size(); i++ ) {
      for ( int j = 0; j < v2.size(); j++ ) {
        if ( v1[i]== v2[j]) {index[j] = 1;}
      }
    }
    //loop throught the index vector and check if we have a sequence of 1s
    for ( int i = 0; i < index.size(); i++ ) {
      if ( index[i] == 1 ) {
        for ( int j = i; j < index.size(); j++ ) {
          if ( index[j] == 1 ) {counter++;}
        }
        //if the sequence of 1s = to the size of v1 it means that we have identified the sub-vector
        if(counter == v1.size()){break;}
        else{counter = 0; continue;}
      }
    }
  }//end if
    return counter/(float)v1.size();
}


int main()
{
    std::vector<string> v1{"A","B","C"};
    std::vector<string> v2{"X","A","B","C","Y"};
    cout <<  wordOrder (v1, v2 ) << endl;
    return 0;
}

推荐答案

图书馆。使用 std :: search 执行范围搜索：

vector<string> v1 = {"A","B","C"};
vector<string> v2 = {"X","Y","A","B","C","D"};

auto res = search(begin(v2), end(v2), begin(v1), end(v1));

并测试是否找到范围：

auto found = res != end(v2);

Live example 此处。

Live example here.

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