[Leetcode 692/347]前K个高频词Top K Frequent Words Element

 

题目347

 

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

347简单些，692是其变式，多了一种排序

Given an integer array nums and an integer k, return the k most frequent elements. You may return the answer in any order.

 

 

 

Example 1:

 

Input: nums = [1,1,1,2,2,3], k = 2
Output: [1,2]

 

Example 2:

 

Input: nums = [1], k = 1
Output: [1]

 

 

代码

 

class Solution {
    class Data{
        int num;
        int ncnt;
        Data(int num,int ncnt){
            this.ncnt=ncnt;
            this.num=num;
        }
    }
    public int[] topKFrequent(int[] nums, int k) {
        int[] ans=new int[k];
        Map<Integer,Integer> dataset=new HashMap<>();
        for(int n:nums){
            dataset.put(n,dataset.getOrDefault(n,0)+1);//统计频次
        }

        PriorityQueue<Data> p=new PriorityQueue<Data>(new Comparator<Data>(){
            @Override
            public int compare(Data d1,Data d2){
                return d2.ncnt-d1.ncnt;//降序排列
            }
        });
        
        for(int n:dataset.keySet()){
            p.offer(new Data(n,dataset.get(n)));//插入，进行排序
        }
        for(int i=0;i<k;i++){
            ans[i]=p.poll().num;//取出排序好的数
        }
        return ans;
    }
}

 

题目692

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

数组中有一堆词，求数组中出现频次最高的前K的词

如果出现频次相同，则按字典序abc输出

Given an array of strings words and an integer k, return the k most frequent strings.

Return the answer sorted by the frequency from highest to lowest. Sort the words with the same frequency by their lexicographical order.

 

Example 1:

Input: words = ["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: words = ["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.

 

思路

优先队列，重点是怎么重写comparator，重写格式

        //d1-d2升序排列(abcde，1234)
        //d2-d1降序排列(edcba,4321)
        PriorityQueue<数据结构> q=new PriorityQueue<>(new Comparator<数据结构>() {
            @Override
            public int compare(数据结构 d1, 数据结构d2) {
                return d1-d2;
            }
        });

代码

重新建了个data类，更加直观，也有直接用hashmap的方法

 

class Solution {
    class Data{
        String name;
        int cnt;

        Data(String name,int cnt){
            this.name=name;
            this.cnt=cnt;
        }
    }
    public List<String> topKFrequent(String[] words, int k) {
        List<String> ans = new ArrayList<>();
        Map<String,Integer> dataset=new HashMap<>();
        for(String word:words){
            dataset.put(word,dataset.getOrDefault(word,0)+1);//计算所有word的出现频次
        }

        //d1-d2升序排列(abcde，1234)
        //d2-d1降序排列(edcba,4321)
        PriorityQueue<Data> q=new PriorityQueue<>(new Comparator<Data>() {
            @Override
            public int compare(Data d1, Data d2) {
                if(d1.cnt!=d2.cnt)
                    return d2.cnt-d1.cnt;//频次不等，比较频次
                return d1.name.compareTo(d2.name);//频次相等，比较字母排序
            }
        });

        for(String s:dataset.keySet()){
            q.add(new Data(s,dataset.get(s)));
        }
        for(int i=0;i<k;i++){
            ans.add(q.poll().name);//输出前k个频次高的
        }
        return ans;
    }
}