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:

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.

代碼：

class Solution {
public:
    vector<string> topKFrequent(vector<string>& words, int k) {
        vector<string> res(k);
        unordered_map<string, int> freq;
        auto cmp = [](pair<string, int>& a, pair<string, int>& b) {
            return a.second > b.second || (a.second == b.second && a.first < b.first);
        };
        priority_queue<pair<string, int>, vector<pair<string, int>>, 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 自定義排序 get priority_queue 正常按照第一項從大到小 然後第二項從大到小就不符合題意需要第二項按從小到大 所以自定義

FH 睡著啦

#Leetcode# 692. Top K Frequent Words