[leetcode-347-Top K Frequent Elements]

Given a non-empty array of integers, return the k most frequent elements.

For example,
Given [1,1,1,2,2,3] and k = 2, return [1,2].

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.

思路：

首先用map统计数字出现的次数，然后再将出现的次数作为关键词，使用桶排序，然后从后遍历，返回关键字即可。

vector<int> topKFrequent(vector<int>& nums, int k){if (nums.empty())return{};map<int, int>mp;vector<vector<int>>bucket(nums.size() + 1);for (auto a:nums)mp[a]++;for (auto it : mp)bucket[it.second].push_back(it.first);vector<int>ret;for (int i = nums.size(); i >= 0 && k>0;i--){if (!bucket[i].empty()){for (int j = 0; j < bucket[i].size() && k>0; j++){ret.push_back(bucket[i][j]);k--;}}}return ret;}

后来又看到有人用优先队列存储，感觉更方便了。

 vector<int> topKFrequent(vector<int>& nums, int k) {unordered_map<int,int> map;for(int num : nums){map[num]++;}vector<int> res;// pair<first, second>: first is frequency,  second is numberpriority_queue<pair<int,int>> pq; for(auto it = map.begin(); it != map.end(); it++){pq.push(make_pair(it->second, it->first));if(pq.size() > (int)map.size() - k){res.push_back(pq.top().second);pq.pop();}}return res;}

参考：

https://discuss.leetcode.com/topic/44226/c-o-n-log-n-k-unordered_map-and-priority_queue-maxheap-solution

转载于:https://www.cnblogs.com/hellowooorld/p/7157880.html

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