Delete a variety of methods list list of a certain element of

  When we deal with in business, in many cases data needs to be layers of filtration, recently needed to remove a list of elements does not meet the conditions,

Would have felt quite simple, Google found that many methods are under the old method, did not meet my needs.

So the online reference method to write several methods

1 .Iterator removal

Iterator<AdmVipConfigs> iterator = admVipConfigs.iterator();
while (iterator.hasNext()){
    AdmVipConfigs next = iterator.next();
    String vipName1 = next.getVipName();
    if (vipName1.equals("铂金会员")){
        iterator.remove();
    }
}

Although it can be removed but transformed into a list of the types of Iterator (also converts do not like)

This is not in line with my code style first pass

It has changed the way some older converting find defects are not small (trouble)

This time can no longer follow the old ideas have to go myself to think of a good way of

This time suddenly remembered flow though with not much but it's like this style

Stream how to deal with this problem then

Needless to say definitely to Google

And then found the flow really can handle and more powerful (good-looking)

2 stream to remove

admVipConfigs.stream () findFirst () the Map.. (VO -> { 
    IF (vo.getVipName () the equals ( "registered members")) {. 
        admVipConfigs.remove (VO); 
    } 
    return VO; 
}); 
no change me the list format and more concise, looking quite satisfied 
but when the stream processing remove time will traverse the entire list again to find 
this element, there will be some loss of performance, how it can do it, I'm naked eye can see 
loss rubbish Code (unless no time to change) 

since we can only add to its index of 

3 stream + index removal

IntStream.range (0, admVipConfigs.size ()). Filter (I-> 
    admVipConfigs.get (i) .getVipName (). The equals ( "Gold Membership")). 
Boxed (). FindFirst (). The Map (i -> admVipConfigs.remove ((int) i )); 

which is boxed boxing operation 
seemed simple point, is this the right to be!

 //   remove a certain element in the list 
    @Test
     public  void bb () { 
        String vipName [] = { "registered members", "Gold Member", "Platinum Membership", "Diamond", "supreme members", " super Member " }; 

        List <AdmVipConfigs> = admVipConfigs new new the ArrayList <> ();
         for ( int I =. 1; I <. 7; I ++ ) { 
            AdmVipConfigs admVipConfig = new new AdmVipConfigs (); 
            admVipConfig.setVipId (I); 
            admVipConfig.setVipDel ( 0 ); 
            admVipConfig.setVipDefaultValue ( 0 );
            admVipConfig.setVipOrderNo(i);
            if (i == 1) {
                admVipConfig.setVipStatus(0);
            } else {
                admVipConfig.setVipStatus(1);
            }
            admVipConfig.setVipCondition(0);
            admVipConfig.setVipName(vipName[i - 1]);
            admVipConfigs.add(admVipConfig);
        }
        admVipConfigs.stream().findFirst().map(vo -> {
            if (vo.getVipName().equals("注册会员")){
                admVipConfigs.remove(vo);
            }
            return vo;
        });
        IntStream.range(0,admVipConfigs.size()).filter(i->
            admVipConfigs.get(i).getVipName().equals("金卡会员")).boxed().findFirst().map(i->admVipConfigs.remove((int)i));

        Iterator<AdmVipConfigs> iterator = admVipConfigs.iterator();
        while (iterator.hasNext()){
            AdmVipConfigs next = iterator.next();
            String vipName1 = next.getVipName();
            if (vipName1.equals("铂金会员")){
                iterator.remove();
            }
        }

        System.out.println(admVipConfigs);
    }

The complete code