Question

给定一对具有2个字段x和y的结构向量（在任一向量中找不到重复的x），如何为每个匹配的X对求和每个值Y（或者简单地使用Y表示不匹配X ）是否有捷径可寻？我尝试排序，似乎必须有一种方法来有效地做到这一点，而不使用std :: map

示例：

v1 = [{x = 1，y = 2}，{x = 1000，y = 3}，{x = 3，y = 2}]

v2 = [{x = 0，y = 0}，{x = 1，y = 1}，{x = 3，y = -3}]

PairWiseSum（v1，v2）= [{x = 0，y = 0}，{x = 1，y = 3}，{x = 3，y = -2}，{x = 1000，y = 3 }]

struct mystruct{
   mystruct(int x, double y) {
    X= x;
    Y= y;
  }
  int X;
  double Y;
  bool operator < (const mystruct& other) const
  {
    return (x < other.x);
  }
};

std::vector<mystruct> PairWiseSum(std::vector<mystruct> s1,std::vector<mystruct> s2)
{
   std::vector<mystruct> sumVector;
   sort(s1.begin(), s1.end());
   sort(s2.begin(), s2.end());
   ...
   return sumVector;
}

Answer 1

浏览s1和s2，比较每个集合中的当前项目。如果x值相同，请将它们一起添加。否则，输出mystruct值较小的x。

std::vector<mystruct> PairWiseSum(std::vector<mystruct> s1, std::vector<mystruct> s2)
{
    std::vector<mystruct> sumVector;
    sort(s1.begin(), s1.end());
    sort(s2.begin(), s2.end());

    for (auto current1 = begin(s1), current2 = begin(s2); current1 != end(s1) || current2 != end(s2); )
    {
        if (current1 == end(s1))
            sumVector.push_back(*current2++);
        else if (current2 == end(s2))
            sumVector.push_back(*current1++);
        else if (current1->X < current2->X)
            sumVector.push_back(*current1++);
        else if (current1->X > current2->X)
            sumVector.push_back(*current2++);
        else
        {
            sumVector.emplace_back(current1->X, current1->Y + current2->Y);
            current1++;
            current2++;
        }
    }
    return sumVector;
}

c ++ / c struct array pair wise sum

1 个答案: