比较多类型向量的字符串部分和一种类型向量并删除重复项的更快方法

如何解决比较多类型向量的字符串部分和一种类型向量并删除重复项的更快方法

我想比较下面向量的字符串部分，并尽快删除重复项。

using namespace std;

vector<string> v1;
vector<pair<string,int>> v2;

我尝试了find_if lambda，它似乎比嵌套的for循环快一点。

示例数据：

+--------+------------+
| index  |     v1     |
+--------+------------+
| 0      | apple      |
+--------+------------+
| 1      | watermelon |
+--------+------------+
| 2      | cherry     |
+--------+------------+
| 3      | tomato     |
+--------+------------+
| 4      | cucumber   |
+--------+------------+
| .      | .          |
+--------+------------+
| .      | .          |
+--------+------------+
| 419776 | lettuce    |
+--------+------------+

+--------+---------------------+
| index  |         v2          |
+--------+------------+--------+
|        |    first   | second |
+--------+------------+--------+
| 0      | pear       | 345    |
+--------+------------+--------+
| 1      | apple      | 85     |
+--------+------------+--------+
| 2      | strawberry | 1912   |
+--------+------------+--------+
| 3      | grape      | 54     |
+--------+------------+--------+
| 4      | peach      | 90     |
+--------+------------+--------+
| .      | .          | .      |
+--------+------------+--------+
| .      | .          | .      |
+--------+------------+--------+
| 21803  | pineapple  | 100    |
+--------+------------+--------+

所需结果：

如您所见，apple在两个向量中都完全匹配。所以我想v1删除重复项。

经过测试的方法：

for (auto itr1 = v1.begin(); itr1 != v1.end(); ++itr1)
    for (auto itr2 = v2.begin(); itr2 != v2.end(); ++itr2)
        if (*itr1 == itr2->first)
            v1.erase(itr1);

嵌套for循环花费了大约35秒。

auto itr = v1.begin();
for_each(v1.begin(),v1.end(),[&itr,&v1,v2](const auto& s)
{
    ++itr;
    if (find_if(v2.begin(),v2.end(),[s](const auto& sV) { return s == sV.first; }) != v2.end())
        v1.erase(itr);
});

这个花了大约31秒。

解决方法

#include <iostream>
#include <string>
#include <unordered_set>
#include <vector>
#include <algorithm>
using namespace std;

vector<string> v1{ "apple","watermelon","cherry","cucumber" };
vector<pair<string,int>> v2{ {"pear",345},{"apple",85},{"strawberry",1912},{"grape",54},{"peach",90} };

struct HASH {
public:
    size_t operator()(const pair<string,long long>& p) const {
        return hash<string>()(p.first);
    }
};

struct EQUAL {
public:
    bool operator()(const pair<string,long long>& p1,const pair<string,long long>& p2) const {
        return p1.first == p2.first ? true : false;
    }
};

int main()
{
    vector<pair<string,long long>> V1Unique;
    unordered_set<pair<string,long long>,HASH,EQUAL> s;

    for (auto itr = v2.begin(); itr != v2.end(); ++itr)
        s.insert(make_pair(itr->first,itr-v2.begin()));
    for (auto itr = v1.begin(); itr != v1.end(); ++itr)
        s.insert(make_pair(*itr,v2.size()+itr-v1.begin()));

    V1Unique.assign(s.begin(),s.end());

    sort(V1Unique.begin(),V1Unique.end(),[](const auto& a,const auto& b) { return a.second > b.second; });

    for (int i = 0; i < v2.size(); ++i)
        if (!V1Unique.empty())
            V1Unique.pop_back();

    for (auto itr = V1Unique.begin(); itr != V1Unique.end(); ++itr)
        cout << itr->first << endl;

    return 0;
}

我认为这是最快的方法。