Java 如何从Android中的arraylist中删除重复值
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/1879700/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
How to remove Duplicate value from arraylist in Android
提问by Kumar
ArrayList<String> values=new ArrayList<String>();
values.add("s");
values.add("n");
values.add("a");
values.add("s");
In this Array, I want to remove repeated values.
在这个数组中，我想删除重复的值。
回答by dxh
I think a real neat solution for enforcing unique array lists is this one, if it's not too much code for what you're trying to achieve.
我认为强制执行唯一数组列表的一个真正巧妙的解决方案是这个，如果它不是你想要实现的太多代码。
public class UniqueOverridingList extends ArrayList {
public enum LAST_RESULT {
ADD, OVERRIDE, NOTHING;
}
private LAST_RESULT lastResult;
public boolean add(T obj) {
for (int i = 0; i < size(); i++) {
if (obj.equals(get(i))) {
set(i, obj);
lastResult = LAST_RESULT.OVERRIDE;
return true;
}
}
boolean b = super.add(obj);
if (b) {
lastResult = LAST_RESULT.ADD;
} else {
lastResult = LAST_RESULT.NOTHING;
}
return b;
}
public boolean addAll(Collection c) {
boolean result = true;
for (T t : c) {
if (!add(t)) {
result = false;
}
}
return result;
}
public LAST_RESULT getLastResult() {
return lastResult;
}
}
回答by Scharrels
The class David Hedlund suggested can be made a lot shorter:
David Hedlund 建议的类可以缩短很多：
public class UniqueArrayList extends ArrayList {
/**
* Only add the object if there is not
* another copy of it in the list
*/
public boolean add(T obj) {
if(this.contains(obj))
return false;
return super.add(obj);
}
public boolean addAll(Collection c) {
boolean result = false;
for (T t : c) {
if (add(t)) {
result = true;
}
}
return result;
}
}
The addAlloperation is modified too. The documentationstates:
该addAll操作也被改变。该文件指出：
Returns: true if this list changed as a result of the call.
返回： 如果此列表因调用而更改，则返回 true。
I modified the method to reflect this behaviour. There's still one problem. The documentation of the addAll() method also states:
我修改了方法以反映这种行为。还有一个问题。addAll() 方法的文档还指出：
Appends all of the elements in the specified collection to the end of this list, in the order that they are returned by the specified collection's iterator.
按照指定集合的迭代器返回的顺序，将指定集合中的所有元素追加到此列表的末尾。
The order might be broken by using this method. A possible workaround for this problem might be not supporting the addAllmethod.
使用此方法可能会破坏订单。此问题的可能解决方法可能不支持该addAll方法。
回答by Freyr
Try below code,
试试下面的代码，
ArrayList<String> values=new ArrayList<String>();
String newValue;
// repeated additions:
if (!values.contains(newValue)) {values.add(newValue);}
回答by Vaishali Sutariya
HashSet hs = new HashSet();
hs.addAll(demoArrayList); // demoArrayList= name of arrayList from which u want to remove duplicates
demoArrayList.clear();
demoArrayList.addAll(hs);
回答by Silambarasan Poonguti
Try this...
尝试这个...
ArrayList<String> values=new ArrayList<String>();
HashSet<String> hashSet = new HashSet<String>();
hashSet.addAll(values);
values.clear();
values.addAll(hashSet);
Happy coding...
快乐编码...
