[leetcode]347. Top K Frequent Elements
Analysis
ummmmm~—— [ummmm]
Given a non-empty array of integers, return the k most frequent elements.
先用hash table把每个元素及其出现次数统计出来,然后再根据出现次数排序,输出前k个元素~
Implement
方法一
class Solution {
public:
vector<int> topKFrequent(vector<int>& nums, int k) {
unordered_map<int, int> cnt;
for(auto num:nums)
cnt[num]++;
vector<pair<int, int> > cnt1;
unordered_map<int, int>::iterator it;
for(it=cnt.begin(); it!=cnt.end(); it++){
cnt1.push_back(make_pair(it->first, it->second));
}
sort(cnt1.begin(), cnt1.end(), cmp);
vector<int> res;
vector<pair<int, int> >::iterator it1;
int num=0;
for(it1=cnt1.begin(); it1!=cnt1.end(); it1++){
res.push_back(it1->first);
num++;
if(num == k)
break;
}
return res;
}
static bool cmp(const pair<int, int>& p1, const pair<int, int>& p2){
return p1.second > p2.second;
}
};
方法二(priority queue)
class Solution {
public:
vector<int> topKFrequent(vector<int>& nums, int k) {
unordered_map<int, int> cnt;
for(auto num:nums)
cnt[num]++;
unordered_map<int, int>::iterator it;
priority_queue<pair<int, int> > q;
for(it=cnt.begin(); it!=cnt.end(); it++){
q.push(make_pair(it->second, it->first));
}
vector<int> res;
for(int i=0; i<k; i++){
res.push_back(q.top().second);
q.pop();
}
return res;
}
};