LeetCode:347. Top K Frequent Elements
2016-07-25 13:35
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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.
今天我真是重新见识了一下STL,有一堆容器还没有用过。
最下面用vector的是我写的,当然这种写法也是当年学的别人的,现在快成自己的了–3。
第一个没有注释掉的是我参考了别人写的第二种之后,突发奇想写的。用可以重复,默认升序的map装关键字是出现次数,对应内容数字–1。
中间的是priority_queue,先来个相关说明:
priority_queue:
http://www.cnblogs.com/mfryf/archive/2012/09/05/2671883.html
代码参考:
http://blog.csdn.net/ebowtang/article/details/51317106
默认最大堆。以pair为元素类型,还可以默认排序,估计是key排序。注意在获取最大值后要pop()。
class Solution { public: // static bool compare(pair<int,int> &pa,pair<int,int> &pb) // { // return pa.second>pb.second; // } vector<int> topKFrequent(vector<int>& nums, int k) { unordered_map<int,int> umNums; for(const int &im:nums) { ++umNums[im]; } //1 multimap<int,int> mRes; for(const pair<int,int> &pm:umNums) { mRes.insert({pm.second,pm.first}); } vector<int> res; for(auto iter=mRes.rbegin();iter!=mRes.rend()&&res.size()<k;++iter) { res.push_back(iter->second); } return res; //2 // priority_queue<pair<int,int>> pq; // for(const pair<int,int> &pm:umNums) // { // pq.push({pm.second,pm.first}); // } // vector<int> res(k); // for(int i=0;i<k;++i) // { // res[i]=pq.top().second;//最大堆 // pq.pop(); // } // return res; //3 // vector<pair<int,int>> pvNums;//vector自己加个比较函数排序 // for(const pair<int,int> &pm:umNums) // { // pvNums.push_back(pm); // } // sort(pvNums.begin(),pvNums.end(),compare); // vector<int> res(k); // for(int i=0;i<k;++i) // { // res[i]=pvNums[i].first; // } // return res; } };
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