[leetcode]347. Top K Frequent Elements

[leetcode]347. Top K Frequent Elements

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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;
    }
};