c ++ - 棘手方法 - 需要解决方案 [英] c++ - Tricky Method - need solution

问题描述

对象数组tArray包含买方姓名和购买的数字，每个买方可以在对象数组中多次。我必须在数组中返回五个最大的买家的名字。

The array of objects tArray contains buyer names and the numshares of there purchases, each buyer can be in the array of objects more than once. I have to return in an array the names of the five largest buyers.

我试图与买方名称并行运行两个数组，并且在另一个数组中有总体积。

I attempted to run two arrays in parallel with the buyer name and there total volume in another array.

我的方法一般有缺陷，因为我得到错误的结果，我该如何解决这个问题。

my method in general flawed as i am getting wrong results, how can I solve this problem.

感谢

ntransactions =数组中的事务数

ntransactions = the number of transactions in the array

string* Analyser::topFiveBuyers()
{
//set size and add buyer names for comparison.
const int sSize = 5;
string *calcString = new string[sSize];
calcString[0] = tArray[0].buyerName;
calcString[1] = tArray[1].buyerName;
calcString[2] = tArray[2].buyerName;
calcString[3] = tArray[3].buyerName;
calcString[4] = tArray[4].buyerName;
int calcTotal[sSize] = {INT_MIN, INT_MIN, INT_MIN, INT_MIN, INT_MIN};

//checks transactions
for (int i = 0; i<nTransactions; i++)
{
    //compares with arrays
    for(int j =0; j<sSize; j++)
    {
        //checks if the same buyer and then increase his total
        if(tArray[i].buyerName == calcString[j])
        {
        calcTotal[j] += tArray[i].numShares;
        break;
        }
            //checks if shares is great then current total then replaces
            if(tArray[i].numShares > calcTotal[j])
            {           
            calcTotal[j] = tArray[i].numShares;
            calcString[j] = tArray[i].buyerName;
            break;
            }   
    }
}
return calcString;
}

推荐答案

，我从std :: map开始累加值：

Assuming you're allowed to, I'd start by accumulating the values into an std::map:

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

for (int i=0; i<ntransactions; i++)
    totals[tarray[i].buyername] += tarray[i].numshares;

这将累计每个买家的总股数。然后，您要将该数据复制到std :: vector，并按共享数获取前5名。目前，我将假设你的结构（ buyername 和 numshares 作为成员）被命名为 transaction 。

This will add up the total number of shares for each buyer. Then you want to copy that data to an std::vector, and get the top 5 by number of shares. For the moment, I'm going to assume your struct (with buyername and numshares as members) is named transaction.

std::vector<transaction> top5;

std::copy(totals.begin(), totals.end(), std::back_inserter(top5));

std::nth_element(top5.begin(), top5.begin()+5, top5.end(), by_shares());

为了这个工作，你需要一个比较函数 by_shares 看起来像：

For this to work, you'll need a comparison functor named by_shares that looks something like:

struct by_shares { 
    bool operator()(transaction const &a, transaction const &b) { 
        return b.numshares < a.numshares;
    }
};

或者，如果你使用的编译器足够支持它，你可以使用lambda的比较显式函数：

Or, if you're using a compiler new enough to support it, you could use a lambda instead of an explicit functor for the comparison:

std::nth_element(totals.begin(), totals.end()-5, totals.end(), 
    [](transaction const &a, transaction const &b) { 
        return b.numshares < a.numshares; 
    });

无论哪种方式，在nth_element完成后，你的前5个都会在向量的前5个元素中。我已经扭转了正常的比较这样做，所以它基本上是按降序工作。或者，您可以使用升序，但从集合的末尾指定第5点，而不是从开头指定第5点。

Either way, after nth_element completes, your top 5 will be in the first 5 elements of the vector. I've reversed the normal comparison to do this, so it's basically working in descending order. Alternatively, you could use ascending order, but specify the spot 5 from the end of the collection instead of 5 from the beginning.

我应该补充说，还有其他方法做到这一点 - 例如，一个Boost bimap也会做得很好。鉴于这听起来像家庭作业，我的猜测是一个预打包的解决方案像bimap处理几乎整个工作，你可能不会/不会允许（甚至 std :: map 可能被禁止的原因很相同）。

I should add that there are other ways to do this -- for example, a Boost bimap would do the job pretty nicely as well. Given that this sounds like homework, my guess is that a pre-packaged solution like bimap that handles virtually the entire job for you probably would't/won't be allowed (and even std::map may be prohibited for pretty much the same reason).

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