题解 | #字符串出现次数的TopK问题#

利用PriorityQueue实现一个大顶堆，用map记录好每个字符出现的次数，然后将map的数据导入大顶堆，然后从大顶堆弹出，记录在String[][]，然后返回，，利用大顶堆整理顺序。

import java.util.*;


public class Solution {
    /**
     * return topK string
     * @param strings string字符串一维数组 strings
     * @param k int整型 the k
     * @return string字符串二维数组
     */
    class Tuple implements Comparable<Tuple>{
        String name;
        int index;
        public Tuple(String n,int i){
            this.name = n;
            this.index = i;
        }
        public int compareTo(Tuple t){
            if(this == t){
                return 0;
            }
            if(this.index != t.index){
                return -(this.index - t.index);
            }else{
                return this.name.compareTo(t.name);
            }
        }
    }
    public String[][] topKstrings (String[] strings, int k) {
        // write code here
        String[][] res = new String[k][2];
        if(strings.length == 0 || k == 0){
            return res;
        }
        HashMap<String,Integer> hm = new HashMap<>();
        for(String s : strings){
            if(hm.containsKey(s)){
                int flag = hm.get(s) + 1;
                hm.put(s,flag);
            }else{
                hm.put(s,1);
            }
        }
        PriorityQueue<Tuple> p = new PriorityQueue<>();
        for(HashMap.Entry<String,Integer> h : hm.entrySet()){
            p.offer(new Tuple(h.getKey(),h.getValue()));
        }
        for(int i = 0;i<k;i++){
            Tuple x = p.poll();
            res[i][0] = x.name;
            res[i][1] = String.valueOf(x.index);
        }
        return res;
    }
}