Leetcode 692. Top K Frequent Words 前K个高频单词（最小堆）

692. Top K Frequent Words 前K个高频单词

题目描述

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.

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.

示例:

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.

解答

这道题先扫描一遍数组，记录一下每个单词出现的频率。

然后建立一个最小堆，里面只能存放k个单词，最后将这k个单词安从大到小排列就是前K个高频单词。

代码

class Solution {
public:
 struct comp {
  bool operator()(const pair<string, int> &lhs, const pair<string, int> & rhs) const {
   if (lhs.second == rhs.second) {
    return lhs.first < rhs.first;
   }
   return lhs.second > rhs.second;
  }
 };
    vector<string> topKFrequent(vector<string>& words, int k) {
        unordered_map <string, int> ump;
        for (auto & w : words) {
         ump[w] += 1;
        }

        priority_queue<pair<string, int>, vector<pair<string, int>>, comp> pq;
        for (auto & u : ump) {
         pq.emplace(u.first, u.second);
         if (pq.size() > k) pq.pop();
        }
        vector<string> ans;
        while(!pq.empty()) {
         ans.insert(ans.begin(), pq.top().first);
         pq.pop();
        }
        return ans;
    }

};