从ArrayList中删除所有出现的元素

I am using java.util.ArrayList, I want to remove all the occurrences of a particular element.

我正在使用java.util.ArrayList，我想删除特定元素的所有出现。

    List<String> l = new ArrayList<String>();
    l.add("first");
    l.add("first");
    l.add("second");

    l.remove("first");

It's removing only the first occurrence. But I want all the occurrences to be removed after l.remove("first"); I expect list to be left out only with the value "second". I found by googling that it can be achieved by creating new list and calling list.removeAll(newList). But is it possible to remove all occurrences without creating new list or is there any API available to achieve it ?

它只删除第一次出现。但我希望在l.remove（“first”）之后删除所有出现的事件;我希望列表只剩下值“second”。我通过谷歌搜索发现它可以通过创建新列表并调用list.removeAll（newList）来实现。但是，是否可以在不创建新列表的情况下删除所有实例，或者是否有可用的API来实现它？

5 个解决方案

#1

109

l.removeAll(Collections.singleton("first"));

#2

Another way using Java 8:

使用Java 8的另一种方法：

l.removeIf("first"::equals);

#3

while(l.remove("first")) { }

This removes all elements "first" from the list.

这将从列表中删除“first”的所有元素。

#4

You can use the removeAll() method.

您可以使用removeAll（）方法。

list.removeAll(Arrays.asList("someDuplicateString"));

#5

Since in your example you are using Strings I guess did should do the trick.

因为在你的例子中你使用字符串，我想应该做的伎俩。

for(int i = 0; i < list.size();i++){
    if(list.get(i).equals(someStringNameOrValue)){
        list.remove(i--);
    }
}

Looks like I misunderstood your question. I updated my answer. Am I right? you want to remove all occurrences of "first" ?

看起来我误解了你的问题。我更新了我的答案。我对吗？你想删除所有出现的“第一”？

#1

109

l.removeAll(Collections.singleton("first"));

#2