leetcode 692. Top K Frequent Words-优快云博客

本文介绍了一种高效算法，用于找出列表中出现频率最高的前K个单词，并按频率从高到低排序。文章详细解释了如何使用优先队列实现此功能，并提供了完整的代码示例。

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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.

题目大意：求出现次数前K的单词集合。思路：统计每个单词的数量，然后放到优先队列中，取出前k个就行了
这题主要是熟悉优先队列的用法。其中比较函数cmp可以写成结构体、类、或者lambda表达式.具体见代码

class cmp1{
public:
    bool operator()(pair<int,string> a, pair<int,string> b) {
        if (a.first == b.first) return a.second > b.second; 
        return a.first < b.first;
    }
};
class Solution {
public:
    vector<string> topKFrequent(vector<string>& words, int k) {
        map<string, int> mp;
        for (int i = 0; i < words.size(); ++i) {
            mp[words[i]]++;
        }
        vector<string> v;
        
        auto cmp = [](pair<int, string>& a, pair<int, string>& b) {
            return a.first < b.first || (a.first == b.first && a.second > b.second);
        };
        
        priority_queue<pair<int, string>, vector<pair<int, string> >,decltype(cmp) >q(cmp);
        int cnt = 0;
        for (auto x: mp) {
            q.push({x.second, x.first});
        }
        while(!q.empty()) {
            pair<int, string> pa = q.top();
            v.push_back(pa.second);
            if (v.size() == k) break;
            q.pop();
        }
        return v;
    }
};

转载于:https://www.cnblogs.com/pk28/p/7705596.html