小顶堆
保持堆的大小不超过k+1即可;
时间复杂度方面,本题应该漏了字符串的长度应该是一个常数,否则复杂度不是
class Solution { public: struct Node{ string s; int cnt; Node(string _s, int _cnt):s(_s), cnt(_cnt){} bool operator < (const Node& other) const{ if(cnt == other.cnt) return s < other.s; else return cnt > other.cnt; } }; /** * return topK string * @param strings string字符串vector strings * @param k int整型 the k * @return string字符串vector<vector<>> */ vector<vector<string> > topKstrings(vector<string>& strings, int k) { // write code here unordered_map<string, int> ump; for(string s:strings) ump[s] ++; priority_queue<Node> pq; //小顶堆 for(pair<string,int> p:ump){ pq.push(Node(p.first, p.second)); if(pq.size() > k) pq.pop(); } vector<vector<string>> res; while(!pq.empty()){ Node node = pq.top(); pq.pop(); res.push_back({node.s, to_string(node.cnt)}); } reverse(res.begin(), res.end()); return res; } };