c++ priority_queue應用(重要)
自定義排序
重寫仿函式
struct cmp{ bool operator() ( Node a, Node b ){//預設是less函式 //返回true時,a的優先順序低於b的優先順序(a排在b的後面) if( a.x== b.x ) return a.y> b.y; return a.x> b.x; } };
struct cmp1{
bool operator () ( int a , int b ){
return a > b;
}
};
struct cmp2{
bool operator ()( int s ,int d ){
return s<d;
}
};
priority_queue<int> q;//預設是從大到小。大頂堆
priority_queue<int, vector<int> ,less<int> >q;//從大到小排序。大頂堆
priority_queue<int, vector<int>, greater<int> >q;//從小到大排序。小頂堆
priority_queue < int , vector<int> , cmp2 > q;//從大到小。大頂堆
priority_queue < int , vector<int> , cmp1 > q;//從小到大。大頂堆
關於priority_queue中元素的比較
模板申明帶3個引數:priority_queue<Type, Container, Functional>,其中Type 為資料型別,Container為儲存資料的容器,Functional 為元素比較方式。
Container必須是用陣列實現的容器,比如vector,deque等等,但不能用 list。STL裡面預設用的是vector。
如果把後面2個引數預設的話,優先佇列就是大頂堆(降序)
347.Top K Frequent Elements
Given a non-empty array of integers, return thekmost frequent elements.
Example 1:
Input: nums = [1,1,1,2,2,3], k = 2
Output: [1,2]
Example 2:
Input: nums = [1], k = 1
Output: [1]
Note:
- You may assumekis always valid, 1 ≤k≤ number of unique elements.
- Your algorithm's time complexitymust bebetter than O(nlogn), wherenis the array's size.
- It's guaranteed that the answer is unique, in other words the set of the top k frequent elements is unique.
- You can return the answer in any order
class Solution { public: struct cmp{ bool operator()(pair<int,int>& a,pair<int,int>& b){ if(a.second >= b.second) return true; else return false; } }; vector<int> topKFrequent(vector<int>& nums, int k) { //sort(nums.begin(),nums.end()); vector<int> res; map<int,int> freq2num; for(int i=0;i<nums.size();i++){ freq2num[nums[i]]++; } //預設大頂堆,我們需要小頂堆 priority_queue<pair<int,int>,vector<pair<int,int>>,cmp> p; for(auto iter=freq2num.begin();iter!=freq2num.end();iter++){ if(p.size()<k){ p.push(*iter); }else{ if(p.top().second<iter->second){ p.pop(); p.push(*iter); } } } while(p.size()){ res.push_back(p.top().first); p.pop(); } return res; } };
vector<int> topKFrequent(vector<int>& nums, int k) { vector<int> res; map<int,int> m; for(auto c:nums) { m[c]++; }
//預設大頂堆,我們選前k個。 priority_queue<pair<int,int>> q; //map中iter->first是要返回的數字,ietr->second是數字的個數 for(auto iter=m.begin();iter!=m.end();iter++) { pair<int,int> pr=make_pair(iter->second,iter->first); q.push(pr); } while(k--) { res.push_back(q.top().second); q.pop(); } return res; }