[Leetcode] 692. Top K Frequent Words 解题报告

本文介绍了一种高效查找列表中出现频率最高的k个单词的算法。该算法首先使用哈希表统计每个单词出现的次数，然后利用大顶堆来保存这些单词及其对应的频率，最后输出前k个最频繁出现的单词。此方法能在O(n log k)时间内完成任务，适用于需要快速处理大量文本数据的场景。

题目：

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.

思路：

分为三步走：1）定义哈希表hash，建立从单词到其出现频次的映射；2）根据哈希表建立一个大顶堆heap；3）依次取出大顶堆的前k个元素，并且返回。

算法的时间复杂度是O(2n + klogn)，空间复杂度是O(n)。

代码：

class Solution {
public:
    vector<string> topKFrequent(vector<string>& words, int k) {
        // step 1: calculate the frequency of each word
        unordered_map<string, int> hash;
        for (auto &word : words) {
            ++hash[word];
        }
        // step 2: construct the heap
        vector<pair<int, string>> heap;
        for (auto it = hash.begin(); it != hash.end(); ++it) {
            heap.push_back(make_pair(it->second, it->first));
        }
        make_heap(heap.begin(), heap.end(), pairCompare);
        // step 3: construct the result
        vector<string> ans;
        for (int i = 0; i < k; ++i) {
            ans.push_back(heap[0].second);
            pop_heap(heap.begin(), heap.end(), pairCompare);
            heap.pop_back();
        }
        return ans;
    }
private:
    static bool pairCompare(pair<int, string> &a, pair<int, string> &b) {
        if (a.first == b.first) {
            return a.second > b.second;
        }
        else {
            return a.first < b.first;
        }
    }
};