Leetcode 刷题 - 排序（day3）

Input:
“tree”
Output:
“eert”
Explanation:
‘e’ appears twice while ‘r’ and ‘t’ both appear once.
So ‘e’ must appear before both ‘r’ and ‘t’. Therefore “eetr” is also a valid answer.

1.建立map字典，统计字符串中每个元素出现的次数2.建立集合数组，以map中的值为集合数组下表，map的键为集合数组元素3.将集合数组从大到小赋值给新集合。因为可能r数组中的某些值为null，故要进行该操作 字符串方法简介

package order;



import java.lang.reflect.Array;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.stream.Stream;


public class Order3_frequencySort {
    public static void main(String[] args) {
        String str = "tree";
        char[] ch = str.toCharArray();
        //1.建立map字典，统计字符串中每个元素出现的次数
        Map frequencyForNum = new HashMap();
        for (char c: ch){
            //关键一步，主要功能就是统计字符串中出现相同字符的数目
            frequencyForNum.put(c,frequencyForNum.getOrDefault(c,0)+1);
        }
        //2.建立集合数组，以map中的值为集合数组下表，map的键为集合数组元素
        List[] frequencyBucket = new ArrayList[ch.length+1];
        for (char c: frequencyForNum.keySet()) {
            int f  = frequencyForNum.get(c);
            if (frequencyBucket[f]==null){
                frequencyBucket[f] = new ArrayList<>();
            }
            frequencyBucket[f].add(c);
        }

        //3.将集合数组从大到小赋值给新集合。因为可能r数组中的某些值为null，故要进行该操作
        
        StringBuilder stringBuilder = new StringBuilder();
        StringBuffer stringBuffer = new StringBuffer();
        List results = new ArrayList<>();
        for (int i = frequencyBucket.length-1;i >= 0;i--){
            if (frequencyBucket[i] == null)
                continue;
            for (char c: frequencyBucket[i]){
                for (int j=0;j 
输出结果