为什么无序设置混合值

std :: unordered_set：

内部，元素未按任何特定顺序排序，而是 组织成水桶。将元素放入哪个水桶的依赖 完全取决于其价值。这允许快速访问 单个元素，由于一旦计算了哈希，它就指的是 确切的水桶将元件放入。

如果您需要订购唯一的终端，请使用std :: set：

#include <iostream>
#include <vector>
#include <string>
#include <set>
int main() {
    std::vector<std::string>terminals = { "e", "d", "a", "b", "c", "a", "a", "a", "a", "b","b", "b", "b", "c", "z" };
    for(const std::string& terminal : terminals) // print given vector
        std::cout << terminal << " ";
    std::cout << "n";;
    // populate the set directly from the vectors iterators:
    std::set<std::string> newSet(terminals.begin(), terminals.end());;
    std::cout << "printing the (ordered) set:" << "n";;
    for(const std::string& terminal : newSet)
        std::cout << terminal << " ";
    std::cout << "n";;
}

如果要维护原始订单，则不能将任一设置用作最终存储您的最终存储。

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <unordered_set>
int main() {
    std::vector<std::string>terminals = { "e", "d", "a", "b", "c", "a", "a", "a", "a", "b","b", "b", "b", "c", "z" };
    for(const std::string& terminal : terminals) // print given vector
        std::cout << terminal << " ";
    std::cout << "n";;
    std::vector<std::string> newSet; // not really a set anymore
    std::unordered_set<std::string> cache; // blacklist
    // try to insert all terminals and only when an insert is successful,
    // put the terminal in newSet
    std::for_each(terminals.begin(), terminals.end(),
        [&](const std::string& terminal) {
            auto [it, inserted] = cache.insert(terminal);
            if(inserted)
                newSet.push_back(terminal);
        }
    );
    std::cout << "printing the vector of unique terminals:" << "n";;
    for(const std::string& terminal : newSet)
        std::cout << terminal << " ";
    std::cout << "n";;
}

如果您想要原始顺序和不介意直接对原始terminals向量进行更改，则可以使用std::remove_if与unordered_set合并，这很不错，因为它不需要新的向量。这是@marek r答案的注释变体：

首先阅读：擦除 - 删除成语

int main() {
    std::vector<std::string>terminals = { "e", "d", "a", "b", "c", "a", "a", "a", "a", "b","b", "b", "b", "c", "z" };
    for(const std::string& terminal : terminals) // print given vector
        std::cout << terminal << " ";
    std::cout << "n";;
    std::unordered_set<std::string> cache; // blacklist
    // remove_if() moves all entries in your container, for which the
    // UnaryPredicate(*) returns true, to the end of the container. It returns
    // an iterator pointing to the first element in the vector that was
    // moved - which is a suitable starting point for a subsequent erase().
    //
    // (*) UnaryPredicate: A callable that returns true or false given a single
    //                     value.
    // auto past_new_end = std::vector<std::string>::iterator past_new_end
    auto past_new_end = std::remove_if(terminals.begin(), terminals.end(),
        // this lambda is the UnaryPredicate
        [&](const std::string& terminal) {
            // insert returns a std::pair<Iterator, bool>
            // where the bool (.second in the pair) is false
            // if the value was not inserted (=it was already present)
            return cache.insert(terminal).second == false;
        }
    );
    std::cout << "display all the entries (now with unspecified values) "
                 "that will be erased:n";
    std::copy(past_new_end, terminals.end(),
                            std::ostream_iterator<std::string>(std::cout, "<"));
    std::cout << "n";
    // erase all the moved entries
    terminals.erase(past_new_end, terminals.end());
    std::cout << "printing the unique terminals:" << "n";;
    for(const std::string& terminal : terminals)
        std::cout << terminal << " ";
    std::cout << "n";;
}

如果要维护原始订单，但是执行唯一性，则可能要：

1. 在项目中阅读。
2. 尝试将其插入集合
3. 如果成功了，则不在集合中，因此也将其复制到输出
4. 重复

如果您要排序的输出（在您的示例中，输出将是" a b c d e z"），那么您可以将项目插入std::set中，否则您可以使用std::sort，然后使用std::unique，然后获得CC_7输入中的每个唯一元素。

看起来您想使用（有序）集。

编辑：实际上看起来不像。std::vector可以起作用，但它可能不是最干净的解决方法。

您也可以使用无序的地图，然后将项目作为地图的键，索引作为对该密钥的相应值。

我正在尝试使用unordered_set从矢量中删除重复项。

为什么您认为unordered_set可以保守任何形式的订单？名称清楚地指出没有任何特定顺序。

您应使用std::unordered_set0仅在序列中找到项目。基于此，您可以从序列中删除项目，因此应该这样做：

void removeDuplicates(Data &data)
{
    std::unordered_set<std::string> foundItems;
    auto newEnd = std::remove_if(data.begin(), data.end(), [&foundItems](const auto &s)
                                 {
                                     return !foundItems.insert(s).second;
                                 });
    data.erase(newEnd, data.end());
}

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