LCR 060. 前 K 个高频元素

优先队列轻松解决

struct node{int num;int cnt;node(int n){num = n;cnt = 0;}bool operator < (const  node m) const{return cnt > m.cnt;}
};class Solution {
public:vector<int> topKFrequent(vector<int>& nums, int k) {unordered_map<int,node*> mp;vector<node> vec;for(int i=0;i<nums.size();++i){if(mp.find(nums[i]) == mp.end()){node* temp = new node(nums[i]);mp[nums[i]] = temp; }mp[nums[i]]->cnt++;}for(auto [u,v] : mp){vec.push_back(*v);}sort(vec.begin(),vec.end());vector<int> ans;for(int i=0;i<k;++i){ans.push_back(vec[i].num);}return ans; }
};

使用拉姆达公式更为简介

class Solution {
public:vector<int> topKFrequent(vector<int>& nums, int k) {unordered_map<int,int> mp;for(auto num : nums){mp[num]++;}auto cmp = [](pair<int,int> &a, pair<int,int> &b){return a.second > b.second ;} ;priority_queue< pair<int,int> ,vector<pair<int,int>> ,decltype(cmp) > minHeap(cmp);for(auto [u,v] : mp ){minHeap.push({u,v});if(minHeap.size() > k){minHeap.pop();}}vector<int> ans; while(!minHeap.empty()){ans.push_back(minHeap.top().first);minHeap.pop();}return ans;}
};