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.

主要分两步 : 第一步就是用unordered_map统计各个数字的次数，key为数字，value为次数。第二步就是针对次数进行排序。因为map默认是对key值升序排序，所以本题重点是第二步。第二步不能再构造key为次数，value为次数的map，因为不同数的次数可能一样，而map中key值是独一性的；所以为了方便，方法一使用sort，方法二使用priority_queue。当然map也可以按方法一sort那样自己重载排序规则。

bool compare(pair<int,int> &a, pair<int,int> &b){
return a.second>b.second;
}

class Solution {
public:
vector<int> topKFrequent(vector<int>& nums, int k) {
unordered_map<int,int> cnt;
vector<int> res;
for(auto &x:nums) cnt[x]++;
vector<pair<int,int> > hist;
for (auto &x:cnt){
hist.push_back(make_pair(x.first,x.second));
}
sort(hist.begin(),hist.end(),compare);
for (int i=0;i<k;i++){
res.push_back(hist[i].first);
}
return res;
}
};

class Solution {
public:
vector<int> topKFrequent(vector<int>& nums, int k) {
unordered_map<int,int> map;
for(int num : nums)
map[num]++;
vector<int> res;
priority_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;
}
};