题解 | #字符统计#

我只会笨办法，先计算数量，根据数量重新拼接字符，当有相同数量的字符时，转成char数组，再排序，最后再拼接所有的输出。

public class Main {    
	public static void main(String[] args) {       
	  Scanner in = new Scanner(System.in);        
	  while (in.hasNextLine()) {            
		String str = in.nextLine();            
		if (str.length() < 1 || str.length() > 1000) {               
			System.out.println("ERROR");               
			continue;           
		}            
		char[] chs = str.toCharArray();           
		Map<String, Integer> countMap = new HashMap<>();            
		for (int i = 0; i < chs.length; i++) {                
		  String sch = "" + chs[i];                
		  Integer ioch = countMap.get(sch);             
		  if (ioch == null) {                
		  countMap.put(sch, 1);              
		} else {                    
		  countMap.put(sch, ioch + 1);        
		  }           
		}         
		List<Integer> countList = new ArrayList<>(countMap.values());   
	  	Collections.sort(countList);            
		// 去重            
		countList = countList.stream().distinct().collect(Collectors.toList());         
		StringBuilder sb = new StringBuilder();          
		for (int i = countList.size() - 1; i >= 0; i--) {          
		  int strLen = countList.get(i);                
		  // 拼接重复数量的字符，进行排序               
		  StringBuilder sortSb = new StringBuilder();         
		  for (Map.Entry<String, Integer> entry : countMap.entrySet()) {         
			  String chi = entry.getKey();           
			  Integer chc = entry.getValue();               
			  if (chc == strLen) {                        
			  		sortSb.append(chi);                   
		  		}              
			}              
			char[] sortCh = sortSb.toString().toCharArray();            
			Arrays.sort(sortCh);             
			sb.append(new String(sortCh));         
			}          
			System.out.println(sb);    
	  	}  
	}
}