66 lines
1.9 KiB
C++
66 lines
1.9 KiB
C++
// Source : https://leetcode.com/problems/top-k-frequent-elements/
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// Author : Calinescu Valentin
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// Date : 2016-05-02
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/***************************************************************************************
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*
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* Given a non-empty array of integers, return the k most frequent elements.
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*
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* For example,
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* Given [1,1,1,2,2,3] and k = 2, return [1,2].
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*
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* Note:
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* You may assume k is always valid, 1 ≤ k ≤ number of unique elements.
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* Your algorithm's time complexity must be better than O(n log n), where n is the
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* array's size.
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*
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***************************************************************************************/
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class Solution {
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public:
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struct element//structure consisting of every distinct number in the vector,
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//along with its frequency
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{
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int number, frequency;
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bool operator < (const element arg) const
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{
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return frequency < arg.frequency;
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}
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};
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priority_queue <element> sol;//we use a heap so we have all of the elements sorted
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//by their frequency
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vector <int> solution;
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vector<int> topKFrequent(vector<int>& nums, int k) {
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sort(nums.begin(), nums.end());
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int i = 1;
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for(; i < nums.size(); i++)
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{
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int freq = 1;
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while(i < nums.size() && nums[i] == nums[i - 1])
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{
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i++;
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freq++;
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}
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element el;
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el.number = nums[i - 1];
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el.frequency = freq;
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sol.push(el);
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}
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if(i == nums.size())//if we have 1 distinct element as the last
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{
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element el;
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el.number = nums[nums.size() - 1];
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el.frequency = 1;
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sol.push(el);
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}
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while(k)//we extract the first k elements from the heap
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{
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solution.push_back(sol.top().number);
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sol.pop();
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k--;
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}
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return solution;
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}
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};
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