问题描述

我有一个整数数组，我需要从中删除重复项，同时保持每个整数第一次出现的顺序.我可以看到这样做，但想象有更好的方法可以更好地利用 STL 算法?插入超出了我的控制范围，因此我无法在插入之前检查重复项.

I have an array of integers that I need to remove duplicates from while maintaining the order of the first occurrence of each integer. I can see doing it like this, but imagine there is a better way that makes use of STL algorithms better? The insertion is out of my control, so I cannot check for duplicates before inserting.

int unsortedRemoveDuplicates(std::vector<int> &numbers) {
    std::set<int> uniqueNumbers;
    std::vector<int>::iterator allItr = numbers.begin();
    std::vector<int>::iterator unique = allItr;
    std::vector<int>::iterator endItr = numbers.end();

    for (; allItr != endItr; ++allItr) {
        const bool isUnique = uniqueNumbers.insert(*allItr).second;

        if (isUnique) {
            *unique = *allItr;
            ++unique;
        }
    }

    const int duplicates = endItr - unique;

    numbers.erase(unique, endItr);
    return duplicates;
}

如何使用 STL 算法做到这一点?

How can this be done using STL algorithms?

推荐答案

naive 的方法是使用 std::set 就像每个人都告诉你的那样.它是矫枉过正并且缓存局部性很差(慢).
smart* 方法是适当地使用 std::vector (确保看到底部的脚注):

The naive way is to use std::set as everyone tells you. It's overkill and has poor cache locality (slow).
The smart* way is to use std::vector appropriately (make sure to see footnote at bottom):

#include <algorithm>
#include <vector>
struct target_less
{
    template<class It>
    bool operator()(It const &a, It const &b) const { return *a < *b; }
};
struct target_equal
{
    template<class It>
    bool operator()(It const &a, It const &b) const { return *a == *b; }
};
template<class It> It uniquify(It begin, It const end)
{
    std::vector<It> v;
    v.reserve(static_cast<size_t>(std::distance(begin, end)));
    for (It i = begin; i != end; ++i)
    { v.push_back(i); }
    std::sort(v.begin(), v.end(), target_less());
    v.erase(std::unique(v.begin(), v.end(), target_equal()), v.end());
    std::sort(v.begin(), v.end());
    size_t j = 0;
    for (It i = begin; i != end && j != v.size(); ++i)
    {
        if (i == v[j])
        {
            using std::iter_swap; iter_swap(i, begin);
            ++j;
            ++begin;
        }
    }
    return begin;
}

然后你可以像这样使用它:

Then you can use it like:

int main()
{
    std::vector<int> v;
    v.push_back(6);
    v.push_back(5);
    v.push_back(5);
    v.push_back(8);
    v.push_back(5);
    v.push_back(8);
    v.erase(uniquify(v.begin(), v.end()), v.end());
}

*注意:这是在典型情况下的聪明方法，其中重复的数量不会太高.如需更全面的性能分析，请参阅this related answer to a related question.

*Note: That's the smart way in typical cases, where the number of duplicates isn't too high. For a more thorough performance analysis, see this related answer to a related question.

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