remove_if等价于std :: map

问题描述

我正在试图从特定情况下从地图中删除一系列元素。如何使用STL算法？

I was trying to erase a range of elements from map based on particular condition. How do I do it using STL algorithms?

最初我想到使用 remove_if ，但不可能的是remove_if对于关联容器不起作用。

Initially I thought of using remove_if but it is not possible as remove_if does not work for associative container.

有没有可用于map的remove_if等效算法？

Is there any "remove_if" equivalent algorithm which works for map ?

作为一个简单的选择，我想到循环遍历地图并清除。但是循环遍历地图并擦除一个安全的选项？（因为迭代器在擦除后无效）

As a simple option, I thought of looping through the map and erase. But is looping through the map and erasing a safe option?(as iterators get invalid after erase)

我使用以下示例：

bool predicate(const std::pair<int,std::string>& x)
{
    return x.first > 2;
}

int main(void) 
{

    std::map<int, std::string> aMap;

    aMap[2] = "two";
    aMap[3] = "three";
    aMap[4] = "four";
    aMap[5] = "five";
    aMap[6] = "six";

//      does not work, an error
//  std::remove_if(aMap.begin(), aMap.end(), predicate);

    std::map<int, std::string>::iterator iter = aMap.begin();
    std::map<int, std::string>::iterator endIter = aMap.end();

    for(; iter != endIter; ++iter)
    {
            if(Some Condition)
            {
                            // is it safe ?
                aMap.erase(iter++);
            }
    }

    return 0;
}

推荐答案

几乎。

Almost.

for(; iter != endIter; ) {
            if (Some Condition) {
                    aMap.erase(iter++);
            } else {
                    ++iter;
            }
}

你最初会增加迭代器如果你从它删除一个元素，那么两次您可以跳过需要擦除的元素。

What you had originally would increment the iterator twice if you did erase an element from it; you could potentially skip over elements that needed to be erased.

这是一个常见的算法，我在许多地方使用过并记录在案。

This is a common algorithm I've seen used and documented in many places.

您是正确的，迭代器在擦除后无效，但只有迭代器引用被擦除的元素，其他迭代器仍然有效。因此在erase（）调用中使用iter ++。

You are correct that iterators are invalidated after an erase, but only iterators referencing the element that is erased, other iterators are still valid. Hence using iter++ in the erase() call.

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