c ++如何找到具有给定值最接近值的映射的键

Question

我想找到与排序映射最接近值的键。例如：

#include <iostream>
#include <map>

int main ()
{
    std::map<int,int> mymap;

    mymap[1]=10;
    mymap[2]=40;
    mymap[3]=100;
    mymap[4]=200;
    mymap[5]=500;

    int wantedvalue=50;
    int wantedindex=mymap.whatshouldIdohere(wantedvalue);

    std::cout<<"The closest value of "<<wantedvalue<<" in the map is located";
    std::cout<<" on "<<wantedindex<<" and is "<<mymap[wantedindex]<<std::endl;
    //Should be:
    //The closest value of 50 in the map is located on 2 and is 40

    return 0;
}

评论中提到的代码应该 return 索引，因为想要的值 50 比任何其他值都更接近第二个位置。

有什么方法可以做到这一点吗？

PS：我知道我可以 "for"，搜索整个地图并在发现大于给定值的值时停止，但最糟糕的执行时间是搜索整个地图table。我还需要运行这么多次，所以我正在寻找比这更好的东西。

Answer 1

使用此容器，您只能使用线性搜索。例如，您可以使用标准算法 std::min_element。否则你应该使用另一个容器。

给你

#include <iostream>
#include <map>

int main()
{
    std::map<int, int> mymap;

    mymap[1] = 10;
    mymap[2] = 40;
    mymap[3] = 100;
    mymap[4] = 200;
    mymap[5] = 500;

    int wantedvalue = 50;


    auto it = std::min_element( mymap.begin(), mymap.end(),
        [&](const auto &p1, const auto &p2)
        {
        return
            std::abs(( long )p1.second - wantedvalue) < 
            std::abs(( long )p2.second - wantedvalue);
        });

    std::cout << "The closest value of " << wantedvalue << " in the map is located";
    std::cout << " on " << it->first << " and is " << it->second << std::endl;

    return 0;
}

程序输出为

The closest value of 50 in the map is located on 2 and is 40

c ++如何找到具有给定值最接近值的映射的键

c++ how to find the key of a map that has the closest value of a given value

c++

algorithm

dictionary

find

closest