力扣2300.咒语和药水的成功对数

力扣2300.咒语和药水的成功对数

• 排序 + 二分

  • 对于能整除的数 -1以后一起处理
  • 可以在原数组直接修改

  class Solution {
  public:
      vector<int> successfulPairs(vector<int>& spells, vector<int>& potions, long long success) {
          sort(potions.begin(),potions.end());
          for(int &x:spells)
          {
              long long target = (success-1)/x;
              if(target < potions.back()) 
                  x = potions.end() - ranges::upper_bound(potions,(int)target);
              else 
                  x = 0;
          }
          return spells;
      }
  };

• 也可以另开一个新的

  class Solution {
  public:
      vector<int> successfulPairs(vector<int>& spells, vector<int>& potions, long long success) {
          vector<int> res;
          sort(potions.begin(),potions.end());
          for(int i=0;i<spells.size();i++)
          {
              long long target = (success-1)/spells[i];
              if(target < potions.back()) 
                  res.emplace_back((potions.end() - ranges::upper_bound(potions,(int)target)));
              else 
                  res.emplace_back(0);
          }
          return res;
      }
  };