题解 | #设计LRU缓存结构#

来一个不考虑效率的版本。

class Solution {
public:
    /**
     * lru design
     * @param operators int整型vector<vector<>> the ops
     * @param k int整型 the k
     * @return int整型vector
     */
     int set(int key,int val)
     {
         for(int i=0;i<count_;i++)
         {
             if(v_[i].first==key)
             {
                 v_[i].second=val;
                 return 0;
             }
         }
         if(count_>=size_)
         {
             count_--;
         v_.erase(v_.begin());
         }
         v_.push_back({key,val});
         count_++;
         return 0;
     }
     int get(int key)
     {
       for(int i=0;i<count_;i++)
       {
           if(v_[i].first==key)
           {
              int val1=v_[i].first;
              int val2=v_[i].second;
              v_.erase(v_.begin()+i);
              count_--;
              set(val1,val2);
              return val2;
           }
       }
         return -1;
     }
    vector<int> LRU(vector<vector<int> >& operators, int k) {
        // write code here
        //初始化。
        size_=k;
        count_=0;

        int n=operators.size();
        vector<int> ans;
        for(int i=0;i<n;i++)
        {
            if(operators[i].size()==3)
            {
                set(operators[i][1],operators[i][2]);
            }else
            {
                ans.push_back(get(operators[i][1]));
            }
        }
        return ans;
    }
    private:
    int size_;
    int count_;
    vector<pair<int,int>> v_;
};