如何从数组c ++中获取唯一的字符串

Question

我知道我的问题对某些人来说可能很愚蠢，但我整天都在谷歌上搜索并尝试制定自己的解决方案，但我失败了..请帮忙..

我需要从一个简单的字符串数组中打印所有唯一的字符串。

示例：

输入：“嗨”“我的”“名字”“嗨”“土豆”“文本”“名字”“嗨”

输出：“我的”“土豆”“文本”

我只做一次打印所有内容的函数（“Hi”、“my”、“name”、“potato”、“text”），但我需要忽略数组中 2 倍或更多次的所有内容。

我的算法是： 1. 冒泡排序

1. 使用基本的 for 和 if 仅打印排序序列中的最后一个字符串

.. if(array[i]!=array[i+1]) //做点什么...

Answer 1

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

using namespace std;
int main()
{
    vector<std::string> v = {
        "Hi", "my", "name", "Hi", "potato", "text", "name", "Hi",
    };

    sort(v.begin(), v.end());
    for (auto a = v.begin(), b = a; a != v.end(); a = b) {
        b = find_if(b, v.end(), [&](string s) {return *b != s;});
        if (distance(a, b) == 1)
                cout << *a << '\n';
    }
}

Answer 2

更新：我误解了这个问题，现在它可以作为问题的输出 简单地说，您可以计算每个字符串的出现次数并仅打印出现的字符串。 时间复杂度：O(N^2) 这是代码

#include<iostream>
#include<set>
#include <string>
#include <vector>
using namespace std;
int main()
{
    int n; // number of strings you want in your array
    cin >> n;
    string t; // get t and push back it in the vector
    vector <string> words; //we use vector to store  as we will push back them
    for(size_t i = 1;i <= n;i++)
    {
        cin >> t;
        words.push_back(t);

    }
    for(int i = 0;i < words.size();i++)
    {
        int cnt = 0;
        for(int j = 0;j < words.size() && cnt < 2;j++)
        {
            if(words[i] == words[j])
                cnt++;
        }
        if(cnt == 1) //its unique..print it
            cout << words[i] <<endl;
    }

}

Answer 3

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

int main()
{
    std::vector<std::string> words{"Hi", "my", "name", "Hi", "potato", "text", "name", "Hi"};

    std::sort(words.begin(), words.end());
    for (auto curr = words.begin(); curr != words.end(); ) {
        // If working w/ few duplicate words:
        auto next = std::find_if(
            curr + 1, words.end(), [&](const auto& s) { return s != *curr; }
            );
        /* If working w/ many duplicate words:
        auto next = std::upper_bound(curr + 1, words.end(), *curr); */
        if (std::distance(curr, next) == 1) {
            std::cout << *curr++ << '\n';
        } else {
            curr = next;
        }
    }
}

Demo

---

std::sort、std::upper_bound、std::find_if、std::distance

Answer 4

#include <algorithm>
#include <iostream>
#include <iterator>
#include <string>
#include <vector>

int main()
{
    std::vector<std::string> words{"Hi", "my", "name", "Hi", "potato", "text", "name", "Hi"};
    std::vector<std::string> out;

    std::sort(words.begin(), words.end());
    std::unique_copy(words.begin(), words.end(), std::back_inserter(out));
    std::set_difference(words.begin(), words.end(), out.begin(), out.end(), std::ostream_iterator<std::string>(std::cout, " "));
}

注意：我在手机上和床上写这篇文章时未经过测试。

Answer 5

#include<iostream>
#include<string>
using namespace std;
int main()
{
int i,j;
string ss;
cin>>ss;
 for(i=0;i<ss.length();i++)
{
    for(j=1;j<ss.length();j++)
    {
     if(ss[i]==ss[j]&&i!=j)
       {
        ss[i]=' ';
       }
    }
}
    for(i=0;i<ss.length();i++)
{
    if(ss[i]!=' ')cout<<ss[i]<<" ";  /// for unique element print
}
return 0;

}

Answer 6

#include<iostream>
using namespace std;
int main()
{
int n;
cin>>n;
int a[n],i,j,k;
for(i=0;i<n;i++)
{
    cin>>a[i];
}
for(i=0;i<n;i++)
{
    for(j=1;j<n;j++)
    {
     if(a[i]==a[j]&&i!=j)
       {
        a[i]=0;
       }
    }
}
for(i=0;i<n;i++)
{
    if(a[i]!=0)cout<<a[i]<<" ";
}

}