对无顺序的map进行排序（升序）以及随机数

/**    
  * @Title: sort @Description: 类型任务数量是否没有达到上限的适用用户，从小到大排序 @param @param   
  *         Map<UserEntity, Long> oldMap @param @return @return String   
  *         userId @throws   
  */   
 public String sortMap(Map<String, Long> oldMap) {   
  // 对无顺序的map进行排序  
  ArrayList<Map.Entry<String, Long>> list = new ArrayList<Map.Entry<String, Long>>(oldMap.entrySet());  
    
  Collections.sort(list, new Comparator<Map.Entry<String, Long>>() {  
   @Override 
   public int compare(Entry<String, Long> arg0, Entry<String, Long> arg1) { 
    return (int) (arg0.getValue() - arg1.getValue());
   } 
  });  
  // 对有序数组进行判断，该类型任务最少的用户为多人的场合，随机分配  
  long lastUserTaskCount = 0;  
  String theSuitableUserId = "";  
  List<String> listSuitableUser = new ArrayList<String>();  
  for (int m = 0; m < list.size(); m++) {  
   long nowUserTaskCount = list.get(m).getValue(); 
   if (m == 0 || lastUserTaskCount == nowUserTaskCount) 
    listSuitableUser.add(list.get(m).getKey());
   else 
    break;
   lastUserTaskCount = list.get(m).getValue(); 
  }  
    
  if (listSuitableUser.size() == 1)  
   theSuitableUserId = listSuitableUser.get(0); 
  else {  
   Random rand = new Random(); 
   int k = rand.nextInt(listSuitableUser.size()); 
   theSuitableUserId = listSuitableUser.get(k); 
  }  
    
  return theSuitableUserId;  
 }