[LeetCode] (medium) 692. Top K Frequent Words

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博客围绕LeetCode上求k个最频繁单词的问题展开，给定非空单词列表，需按频率从高到低返回k个元素，频率相同时按字母顺序排序。还提到假设k有效，输入仅含小写字母，后续可尝试在O(n log k)时间和O(n)额外空间内解决，此外还提及计数建堆时的报错问题。

https://leetcode.com/problems/top-k-frequent-words/

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 = 2
Output: ["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 = 4
Output: ["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:

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

Follow up:

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

计数，建堆：

class Solution {
public:
    class myCompare{
        public:
        myCompare(){};
        bool operator()(pair<string, int> &p1, pair<string, int> &p2) const { 
            if(p1.second == p2.second) return p1.first < p2.first;
            else return p1.second > p2.second;
        }
    };
    
    vector<string> topKFrequent(vector<string>& words, int k) {
        unordered_map<string, int> M;
        for(string str : words){
            ++M[str];
        }
        // priority_queue<pair<string, int>> PQ([](pair<string, int> &p1, pair<string, int> &p2) -> bool {
        //     if(p1.second == p2.second) return p1.first < p2.first;
        //     else return p1.second > p2.second;
        // });
        
        priority_queue<pair<string, int>, vector<pair<string, int>>, myCompare> PQ;
        // priority_queue<pair<string, int>, vector<pair<string, int>>, myCompare> PQ();
        
        for(auto itr = M.begin(); itr != M.end(); ++itr){
            if(PQ.size() < k){
                PQ.emplace(itr->first, itr->second);
            }else{
                if(itr->second > PQ.top().second || (itr->second == PQ.top().second && itr->first < PQ.top().first)){
                    PQ.pop();
                    PQ.emplace(itr->first, itr->second);
                }
            }
        }
        
        vector<string> result(k);
        for(int i = k-1; i >= 0; --i){
            result[i] = PQ.top().first;
            PQ.pop();
        }
        
        return result;
    }
};

*问题：使用第二种声明方式会在PQ.size()处报错：

https://blog.youkuaiyun.com/MaryChow/article/details/69673500