题解 | #字符统计#

java版本，关键在于重写排序函数的比较方式，整体难度一般

```import java.util.*;

public class Main{
    public static int maxn=10001;
   
    public static void main(String[] args){
        class Sub{
            int cnt;
            Character po;
            public void setcnt(int cnt){
                this.cnt=cnt;
            }
            public int getcnt(){
                return this.cnt;
            }
            public void setpo(Character po){
                this.po=po;
            }
            public Character getpo(){
                return this.po;
            }
        }
        Scanner sa=new Scanner(System.in);
        while(sa.hasNext()){
            String s=sa.nextLine();
            Map<Character,Sub> map=new HashMap<>();
            for(int i=0;i<s.length();i++){
                Sub sub=map.get(s.charAt(i));
                if(null==sub){
                    sub=new Sub();
                    sub.setcnt(1);
                    sub.setpo(s.charAt(i));
                }else{
                    sub.setcnt(sub.getcnt()+1);
                }
                map.put(s.charAt(i),sub);
            }
            List<Sub> list=new ArrayList<>();
            for(Map.Entry<Character,Sub> entry: map.entrySet()){
                list.add(entry.getValue());
            }
            Collections.sort(list,new Comparator<Sub>(){
                @Override
                public int compare(Sub sub1,Sub sub2){
                    if(sub1.cnt-sub2.cnt!=0)
                    return sub1.cnt-sub2.cnt;
                    else return sub2.po-sub1.po;
                }
            });
            for(int i=list.size()-1;i>=0;i--)System.out.print(i==0?list.get(i).po+"\n":list.get(i).po);
        }
    }
}