[leetcode] 347. Top K Frequent Elements

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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.

传说中的哈希表。很多不知道的知识

class Solution {  public:      vector<int> topKFrequent(vector<int>& nums, int k) {          unordered_map<int, int> hash;          priority_queue<pair<int,int>> heap;          vector<int> ret;                    for(int num:nums)               hash[num]++;                    for(auto it:hash)               heap.push(make_pair(it.second, it.first));                    for(int i=0; i<k; ++i) {              ret.push_back(heap.top().second);              heap.pop();          }                    return ret;      }  };

知识点

1.priority_queue的用法，具体百度，竟然能自动排序。

2.unordered_map，第一次见啊，好神奇啊

3.for循环中的int num:nums和aoto it:hash。什么鬼啊。感觉好简单

4.make_pair竟然能合二为一

5.it.first代表的key值，it.second代表的值！我去！

只能说66666

下面是小堆取代大堆

class Solution {      typedef pair<int, int> data;  public:      vector<int> topKFrequent(vector<int>& nums, int k) {          unordered_map<int, int> hash;          priority_queue<data, vector<data>, greater<data>> heap;          vector<int> ret;                    for(int num:nums)               hash[num]++;                    for(auto it:hash) {              heap.push(make_pair(it.second, it.first));              if(heap.size() > k) heap.pop();          }                    while(!heap.empty()) {              ret.push_back(heap.top().second);              heap.pop();          }                    return ret;      }  };

差不多啦。关键就是那些新的知识。真是瞎了我的狗眼。哈希啊。套路深啊

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