如何按价值有效地订购地图？ [英] How can I order a map by value efficiently?

问题描述

考虑一个 std :: map< K，V> 。我想通过一个合适的容器 std :: C< V *> 或 std :: C< V& ;> ，在某种意义上没有值的副本来存储C中的元素。此外，C中的元素必须根据 int f V&）应用于每个元素。尽管我的努力，我找不到一个合适的C和一个足够的有效的方式来构建它。你有什么解决方案吗？

Consider a std::map<K,V>. I want to re-order the map by value profiting by an appropriate container std::C<V*> or std::C<V&>, in a way that no copies of values are done to store the elements in C. Furthermore, elements in C must be sorted according to the result of int f(V&) applied to each element. Despite my efforts I could not find an appropriate C and an enough efficient way to build it. Do you have any solution? A small example would be much appreciated.

推荐答案

看起来很简单。

std::map<K,V> src;
int f(V&) {return 0;}

V* get_second(std::pair<const K,V> &r) {return &(r.second);} //transformation
bool pred(V* l, V* r) { return f(*l)<f(*r); }   //sorting predicate

std::vector<V*> dest(src.size());  //make destination big enough
std::transform(src.begin(), src.end(), dest.begin(), get_second); //transformcopy
std::sort(dest.begin(), dest.end(), pred); //sort

除非你的意思是C应该是另一个地图：

Unless you meant C is supposed to be another map:

std::pair<K,V*> shallow_pair(std::pair<const K,V> &r) 
{return std::pair<K,V*>(r.first, &(r.second));}

std::map<K, V*> dest2;
std::transform(src.begin(), src.end(), 
               std::inserter(dest2,dest2.end()), shallow_pair);

http://ideone.com/bBoXq

这需要上一张地图保持在更长的时间 dest ，并且在 dest 被销毁之前没有删除对。否则 src 将需要保存某种类型的智能指针。

This requires the previous map to remain in scope longer than dest, and have no pairs removed until dest is destructed. Otherwise src will need to have been holding smart pointers of some sort.

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