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:
- You may assume k is always valid, 1 ≤ k ≤ number of unique elements.
- Your algorithm's time complexity must be better than O(n log n), where n is the array's size.
- https://www.cnblogs.com/grandyang/p/5454125.html
- A.使用map、priority_queue(最大堆)。
- Runtime: 20 ms, faster than 81.00% of C++ online submissions for Top K Frequent Elements.
- Memory Usage: 11.3 MB, less than 96.77% of C++ online submissions for Top K Frequent Elements.
- Next challenges:
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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;
}
};