问题描述
我得到的最好的例子是我想根据他们的分数对姓名进行排序.
The best example I've got is that I want to sort Names based on their Score.
vector <string> Names {"Karl", "Martin", "Paul", "Jennie"};
vector <int> Score{45, 5, 14, 24};
因此,如果我将分数排序为 {5, 14, 24, 45},则名称也应根据其分数进行排序.
So if I sort the score to {5, 14, 24, 45}, the names should also be sorted based on their score.
推荐答案
正如其他答案中已经建议的那样:结合每个人的姓名和分数可能是最简单的解决方案.
As already suggested in other answers: Combining the name and the score of each individual is likely the simplest solution.
通常,这可以通过有时称为压缩"操作的方式来实现:将两个向量组合成一对向量 - 以及相应的解压缩".
Generically, this can be achieved with what is sometimes referred to as a "zip" operation: Combining two vectors into a vector of pairs - along with a corresponding "unzip".
一般实现,这可能如下所示:
Implemented generically, this may look as follows:
#include <vector>
#include <string>
#include <algorithm>
#include <iostream>
#include <iterator>
// Fill the zipped vector with pairs consisting of the
// corresponding elements of a and b. (This assumes
// that the vectors have equal length)
template <typename A, typename B>
void zip(
const std::vector<A> &a,
const std::vector<B> &b,
std::vector<std::pair<A,B>> &zipped)
{
for(size_t i=0; i<a.size(); ++i)
{
zipped.push_back(std::make_pair(a[i], b[i]));
}
}
// Write the first and second element of the pairs in
// the given zipped vector into a and b. (This assumes
// that the vectors have equal length)
template <typename A, typename B>
void unzip(
const std::vector<std::pair<A, B>> &zipped,
std::vector<A> &a,
std::vector<B> &b)
{
for(size_t i=0; i<a.size(); i++)
{
a[i] = zipped[i].first;
b[i] = zipped[i].second;
}
}
int main(int argc, char* argv[])
{
std::vector<std::string> names {"Karl", "Martin", "Paul", "Jennie"};
std::vector<int> score {45, 5, 14, 24};
// Zip the vectors together
std::vector<std::pair<std::string,int>> zipped;
zip(names, score, zipped);
// Sort the vector of pairs
std::sort(std::begin(zipped), std::end(zipped),
[&](const auto& a, const auto& b)
{
return a.second > b.second;
});
// Write the sorted pairs back to the original vectors
unzip(zipped, names, score);
for(size_t i=0; i<names.size(); i++)
{
std::cout << names[i] << " : " << score[i] << std::endl;
}
return 0;
}
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