Explanation of Java Dictionary Generation Algorithm

In practical applications, people can use a variety of passwords, but no matter how many, their composition is not far from being composed of printable characters, we can think that

class CreateDic{
 private int BitNum;
 private String Str;
 public void SetBitNum(int num)
 {
 BitNum=num;
 }
 public void SetStr(String str)
 {
 Str=str;
 }
 public int GetBitNum(){
 return BitNum;
 }
 public String GetStr(){
 return Str;
 }
 public List<String> GetDic(){
 int[] tmparray=new int[BitNum];
 List<String> final_list=new ArrayList<String>();
 String result="";
 for(int i=0;i<BitNum;i++)
  tmparray[i]=0;
 int nCount=0;
 while(true)
 {
  result="";
  for(int i=0;i<BitNum;i++)
  {
  result+=Str.charAt(tmparray[i]);
  }
  nCount++;
  System.out.println(result);
  final_list.add(result+"\r\n");
  // Begin to proceed 1 Wheel cycle 
  int length=Str.length();
  int mark=0;
  for(int j=BitNum-1;j>=0;j--)
  {
  if(tmparray[j]==length-1){
   if(j!=0){
   continue;
   }
   else{
   mark=1;
   break;
   }
  }
  else{
   tmparray[j]++;
   for(int k=j+1;k<BitNum;k++)
   {
   tmparray[k]=0;
   }
   break;
  }
  }
  if(mark==1){
  break;
  }
 }
 System.out.println("1 Total number of passwords output: "+nCount);
 return final_list;
 }
}

It can be said that if there are three characters 1 that make up the password, which are "abc", and the password length is 6, then BitNum can be set to 6 and the content of Str is "abc", so that all possible password strings can be obtained as return values

Summarize