力扣2300.咒语和药水的成功对数
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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
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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
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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;}};
