[LeetCode] 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：
使用map 存储计算每个元素的出现次数，将map中的<键，值>放入vector中，使用sort函数降序排列，输出前k个对应的数字

bool  cmp (const pair<int, size_t> &x, const pair<int,  size_t> &y)
{
return x.second > y.second;
}
class Solution {
public:
vector<int> topKFrequent(vector<int>& nums, int k)
{
vector<int> res;
unordered_map<int,  size_t> m;
for(int i = 0; i < nums.size(); i++)
{
++m[nums[i]];
}
vector<pair<int,  size_t>> vec(m.begin(),m.end());
sort(vec.begin(), vec.end(),  cmp);
for(int i = 0; i < k; i++)
{
res.push_back(vec[i].first);
}
return res;
}

};