2016-05-02 23:13:48 +03:00

66 lines
1.9 KiB
C++

// Source : https://leetcode.com/problems/top-k-frequent-elements/
// Author : Calinescu Valentin
// Date : 2016-05-02
/***************************************************************************************
*
* 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.
*
***************************************************************************************/
class Solution {
public:
struct element//structure consisting of every distinct number in the vector,
//along with its frequency
{
int number, frequency;
bool operator < (const element arg) const
{
return frequency < arg.frequency;
}
};
priority_queue <element> sol;//we use a heap so we have all of the elements sorted
//by their frequency
vector <int> solution;
vector<int> topKFrequent(vector<int>& nums, int k) {
sort(nums.begin(), nums.end());
int i = 1;
for(; i < nums.size(); i++)
{
int freq = 1;
while(i < nums.size() && nums[i] == nums[i - 1])
{
i++;
freq++;
}
element el;
el.number = nums[i - 1];
el.frequency = freq;
sol.push(el);
}
if(i == nums.size())//if we have 1 distinct element as the last
{
element el;
el.number = nums[nums.size() - 1];
el.frequency = 1;
sol.push(el);
}
while(k)//we extract the first k elements from the heap
{
solution.push_back(sol.top().number);
sol.pop();
k--;
}
return solution;
}
};