347. Top K Frequent Elements

Aurora20052024-08-10 19:41

Given an integer array nums and an integer k, return the k most frequent elements . You may return the answer in any order.

Example 1:

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Input: nums = [1,1,1,2,2,3], k = 2
Output: [1,2]

Example 2:

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Input: nums = [1], k = 1
Output: [1]

Constraints:

1 <= nums.length <= 105
-104 <= nums[i] <= 104
k is in the range [1, the number of unique elements in the array].
It is guaranteed that the answer is unique.

Follow up: Your algorithm's time complexity must be better than O(n log n), where n is the array's size.

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class Solution {
public:
    class mycomparision{
    public://别忘记加上
        bool operator()(const pair<int,int>&lhs,const pair<int,int>& rhs){//重载（）运算符
            return lhs.second>rhs.second;
        }
    };
    vector<int> topKFrequent(vector<int>& nums, int k) {
        //统计元素出现频率
        unordered_map<int,int>map;
        for(int i=0;i<nums.size();i++){
            map[nums[i]]++;
        }
        //频率排序+定义小顶堆
        priority_queue<pair<int,int>,vector<pair<int,int>>,mycomparision>pri_que;
        //用固定大小为K的小顶堆扫描所有频率的大小
        for(unordered_map<int,int>::iterator it=map.begin();it!=map.end();it++){
            pri_que.push(*it);
            if(pri_que.size()>k){
                pri_que.pop();
            }
        }
        //找出前K个高频元素，由于小顶堆先弹出最小的，所以数组倒序，注意要从k-1开始
        vector<int> result(k);
        for (int i = k - 1; i >= 0; i--) {
            result[i] = pri_que.top().first;
            pri_que.pop();
        }
        return result;
    }
};