如何从 stl 集中获取唯一的值对

发布于 2024-09-17 05:02:10 字数 536 浏览 3 评论 0原文

我有一组 stl 整数，我想迭代所有唯一的整数值对，其中通过唯一性，我认为 val1,val2 和 val2,val1 是相同的，我应该只看到该组合一次。

我在 python 中编写了这个，其中使用列表（簇）的索引：

for i in range(len(clusters) - 1):
    for j in range(i+1,len(clusters)):
       #Do something with clusters[i],clusters[j])

但是如果没有索引，我不确定如何使用 stl 集和迭代器实现相同的效果。我尝试过：

for (set<int>::iterator itr = myset.begin(); itr != myset.end()-1; ++itr) {
    cout << *itr;
}

但这失败了，因为迭代器没有 - 运算符。

我怎样才能实现这一点，或者我必须使用不同的容器？

原文

I have a stl set of integers and I would like to iterate through all unique pairs of integer values, where by uniqueness I consider val1,val2 and val2,val1 to be the same and I should only see that combination once.

I have written this in python where I use the index of a list (clusters):

for i in range(len(clusters) - 1):
    for j in range(i+1,len(clusters)):
       #Do something with clusters[i],clusters[j])

but without an index I am not sure how I can achieve the same thing with a stl set and iterators. I tried out:

for (set<int>::iterator itr = myset.begin(); itr != myset.end()-1; ++itr) {
    cout << *itr;
}

but this fails as an iterator doesn't have a - operator.

How can I achieve this, or must I use a different container?

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分分钟 2024-09-24 05:02:10

遵循以下几行怎么样：

for(set<int>::const_iterator iter1 = myset.begin(); iter1 != myset.end(); ++iter1) {
    for(set<int>::const_iterator iter2 = iter1; ++iter2 != myset.end();) {
    {
        std::cout << *iter1 << " " << *iter2 << "\n"; 
    }
}

这会产生所有 N*(N-1)/2 唯一对，其中 N 是集合中整数的数量。

顺便说一句：每当您迭代容器而不修改任何内容时，请使用 const_iterator ，它是很好的风格并且可能具有更好的性能。

编辑：修改代码以反映 Steve Jessop 提出的建议。

How about something along the following lines:

for(set<int>::const_iterator iter1 = myset.begin(); iter1 != myset.end(); ++iter1) {
    for(set<int>::const_iterator iter2 = iter1; ++iter2 != myset.end();) {
    {
        std::cout << *iter1 << " " << *iter2 << "\n"; 
    }
}

This yields all N*(N-1)/2 unique pairs, where N is the number of integers in your set.

As an aside: use a const_iterator whenever you iterate over a container without modifying anything, it's good style and might have better performance.

EDIT: Modified the code to reflect the suggestion made by Steve Jessop.

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不念旧人 2024-09-24 05:02:10

您不需要执行 end() - 1，因为 end() 是一个指向容器中最后一个元素之后的迭代器。

更正后的代码为：

for (set<int>::iterator itr = myset.begin(); itr != myset.end(); ++itr) {
    for (set<int>::iterator itr2 = itr + 1; itr2 != myset.end(); ++itr2) {
        // Do whatever you want with itr and itr2
    }
}

You don't need to do end() - 1 since end() is an iterator that points after the last element in the container.

The corrected code is:

for (set<int>::iterator itr = myset.begin(); itr != myset.end(); ++itr) {
    for (set<int>::iterator itr2 = itr + 1; itr2 != myset.end(); ++itr2) {
        // Do whatever you want with itr and itr2
    }
}

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