从std :: map值获取密钥的有效方法 [英] efficient way to get key from std::map value

问题描述

我的地图如下:

std::map< std::string ,int> mapobj;
mapobj["one"] = 1;
mapobj["two"] = 2;
mapobj["three"] =3 ;

输入为值时如何获取键

EX:

输入:1

输出:一个

注意:在我的情况下，值是唯一的

推荐答案

一对一映射实际上非常容易，最快的方法是维护两个 映射，一个每个方向.如果不是一对一的，它将变得更加复杂，因为您需要提供一种获取值或键的 collection 的方法，而不是单个方法.幸运的是，您只有一对一的要求.

A one-to-one mapping is actually quite easy, the fastest way to do it is to probably maintain two maps, one for each direction. It becomes more complicated if it's not one-to-one since you'll need to provide a way to get a collection of values or key, rather than a single one. Happily, you only have the one-to-one requirement.

其中一张地图就是您现在拥有的一张，另一张地图会将值映射到给定的键，两者都是:

One of the maps is the one you have now, the other will map the values to a given key, soboth would be:

std::map<std::string, int> forwardmapobj;
std::map<int, std::string> reversemapobj;

，这些将在某种 bidimap 类中维护.

and these would be maintained within a bidimap class of some sort.

每当您插入或删除 bidimap 时，都必须在两个内部地图上执行等效的操作.

Whenever you insert to, or delete from, your bidimap, you have to perform the equivalent operation on both internal maps.

例如，这是一些伪代码.它会维护两个映射，并确保它们与您执行的更改键和值的任何操作保持同步:

For example, here's some pseudo-code. It maintains the two maps and ensures that they'e kept in sync for whatever operations you have that change the keys and values:

class biDiMap:
    map<string, int> forwardMap
    map<int, string> reverseMap

    void add(string key, int val):
        if exists forwardMap[key]: throw exception 'duplicate key'
        if exists reverseMap[val]: throw exception 'duplicate value'
        forwardMapObj[key] = val
        reverseMapObj[val] = key

    void delKey(string key):
        if not exists forwardMap[key]: throw exception 'no such key'
        delete reverseMap[forwardMap[key]]
        delete forwardMap[key]

    void delVal(int val):
        if not exists reverseMap[val]: throw exception 'no such value'
        delete forwardMap[reverseMap[val]]
        delete reverseMap[val]

    int getValFor(string key): return forwardMap[key]

    string getKeyFor(int val): return reverseMap[val]

很显然，您可以添加很多 other 东西，但这应该构成基础.无论如何，在将其转换为C ++类之前，您可能已经做了足够的 工作:-)

Obviously, there's plenty of other stuff you could add but that should form the basis. In any case, you've probably got enough work ahead of you turning that into a C++ class :-)

如果您不想想要推出自己的解决方案，那么Boost会提供一个很好的解决方案，您可以按原样使用它. Boost.Bimap 提供了一个完全模板化的双向地图，您应该可以使用最少的代码来使用它，例如以下完整程序:

If you don't want to roll your own solution, then Boost has a very good one that you can pretty well use as is. Boost.Bimap provides a fully-templated bi-directional map that you should be able to use with minimal code, such as the following complete program:

#include <iostream>
#include <string>
#include <boost/bimap.hpp>

using std::string;
using std::cout;
using std::exception;
using boost::bimap;

int main()
{
    typedef bimap<string, int> SiMap;
    typedef SiMap::value_type SiEntry;

    SiMap bidi;
    bidi.insert(SiEntry("ninety-nine", 99));
    int i = 0;
    for (string str: {"one", "two" , "three", "four", "five", "six"}) {
        bidi.insert(SiEntry(str, ++i));
    }

    cout << "The number of entries is " << bidi.size() << "\n\n";

    for (auto i = 1; i <= 7; i += 3) {
        try {
            cout << "Text for number " << i << " is " << bidi.right.at(i) << "\n";
        } catch (exception &e) {
            cout << "Got exception looking up number " << i << ": " << e.what() << "\n";
        }
    }
    cout << "\n";

    for (auto str: {"five", "ninety-nine", "zero"}) {
        try {
            cout << "Number for text '" << str << "' is " << bidi.left.at(str) << "\n";
        } catch (exception &e) {
            cout << "Got exception looking up text '" << str << "': " << e.what() << "\n";
        }
    }
    cout << "\n";

    return 0;
}

它在数字的文本形式和整数值之间创建双向映射，然后进行一些查找(双向)以表明其有效:

It creates a bi-directional mapping between the textual form of a number and the integral value, then does a few lookups (in both directions) to show that it works:

The number of entries is 7

Text for number 1 is one
Text for number 4 is four
Got exception looking up number 7: bimap<>: invalid key

Number for text 'five' is 5
Number for text 'ninety-nine' is 99
Got exception looking up text 'zero': bimap<>: invalid key

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