【CODE】Top K Frequent Elements

347. Top K Frequent Elements

Medium

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Given a non-empty array of integers, return the k most 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:

class Solution {
public:
    vector<int> topKFrequent(vector<int>& nums, int k) {
        int n=nums.size();
        vector<int> res;
        unordered_map<int,int> mp;
        priority_queue<pair<int,int> > q;
        for(int i=0;i<n;i++) mp[nums[i]]++;
        unordered_map<int,int>::iterator it;
        for(it=mp.begin();it!=mp.end();it++) q.push({it->second,it->first});
        for(int i=0;i<k;i++){
            res.push_back(q.top().second);
            q.pop();
        }
        return res;
    }
};
  • B.桶排序:
  • Runtime: 16 ms, faster than 96.53% of C++ online submissions for Top K Frequent Elements.
  • Memory Usage: 13.1 MB, less than 19.35% of C++ online submissions for Top K Frequent Elements.
class Solution {
public:
    vector<int> topKFrequent(vector<int>& nums, int k) {
        int n=nums.size();
        vector<int> res;
        unordered_map<int,int> mp;
        vector<vector<int> > bucket(nums.size()+1);//最多到bucket[nums.size()]
        for(int i=0;i<n;i++) mp[nums[i]]++;
        unordered_map<int,int>::iterator it;
        for(it=mp.begin();it!=mp.end();it++) bucket[it->second].push_back(it->first);
        for(int i=nums.size();i>=0;i--){
            for(int j=0;j<bucket[i].size();j++){
                res.push_back(bucket[i][j]);
                if(res.size()==k) return res;
            }
        }
        return res;
    }
};
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转载自blog.csdn.net/Li_Jiaqian/article/details/103387852