Given a non-empty list of words, return the k most frequent elements.

Your answer should be sorted by frequency from highest to lowest. If two words have the same frequency, then the word with the lower alphabetical order comes first.

Example 1:

Input: ["i", "love", "leetcode", "i", "love", "coding"], k = 2Output: ["i", "love"]Explanation: "i" and "love" are the two most frequent words.    Note that "i" comes before "love" due to a lower alphabetical order.

 

Example 2:

Input: ["the", "day", "is", "sunny", "the", "the", "the", "sunny", "is", "is"], k = 4Output: ["the", "is", "sunny", "day"]Explanation: "the", "is", "sunny" and "day" are the four most frequent words,    with the number of occurrence being 4, 3, 2 and 1 respectively.

 

Note:

1. You may assume k is always valid, 1 ≤ k ≤ number of unique elements.
2. Input words contain only lowercase letters.

 

Follow up:

1. Try to solve it in O(n log k) time and O(n) extra space.

代码：

class Solution {public:    vector
     
       topKFrequent(vector
      
       & words, int k) {        vector
       
         res(k);        unordered_map
        
          freq;        auto cmp = [](pair
         
          & a, pair
          
           & b) { return a.second > b.second || (a.second == b.second && a.first < b.first); }; priority_queue
           
            
             , vector
             
              
               >, decltype(cmp) > q(cmp); for (auto word : words) ++freq[word]; for (auto f : freq) { q.push(f); if (q.size() > k) q.pop(); } for (int i = res.size() - 1; i >= 0; --i) { res[i] = q.top().first; q.pop(); } return res; }};
              
             
            
           
          
         
        
       
      
     

　　priority_queue 自定义排序 get    priority_queue 正常按照第一项从大到小 然后第二项从大到小就不符合题意需要第二项按从小到大 所以自定义

FH 睡着啦