blocks|key|4461414|text|您还可以在Java8中使用filter|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4461415|e.stream().filter(s+->+Collections.frequency(e,+s)+==+1).collect(Collectors.toList())|code-block|syntax|javascript|4461416|entityMap^0|D|6|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

You can also use <code>filter</code> in Java 8

<pre><code>e.stream().filter(s -&gt; Collections.frequency(e, s) == 1).collect(Collectors.toList())
</code></pre>

blocks|key|4461392|text|你可以使用HashMap<String,+Integer>。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4461393|遍历列表，如果Hash+map不包含该字符串，则将其与值1相加。|4461394|另一方面，如果你已经有了字符串，你只需增加计数器即可。因此，字符串的映射将如下所示：|4461395|{"123",+2}
{"122",+1}
{"125",+1}|code-block|syntax|javascript|4461396|然后，您将创建一个新列表，其中每个键的值都是1。|4461397|entityMap^0|5|O|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|V|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|W|8|@]|D|@]|E|$]]|$1|J|3|K|5|L|7|X|8|@]|D|@]|E|$M|N]]|$1|O|3|P|5|6|7|Y|8|@]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|R|$]]

You could use a <code>HashMap&lt;String, Integer&gt;</code>.

You iterate over the list and if the Hash map does not contain the string, you add it together with a value of 1.

If, on the other hand you already have the string, you simply increment the counter. Thus, the map for your string would look like this:

<pre><code>{"123", 2}
{"122", 1}
{"125", 1}
</code></pre>

You would then create a new list where the value for each key is 1.

blocks|key|4461430|text|下面是一个非Java+8的解决方案，它使用map来计算事件的发生次数：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4461431|Map+<String,Integer>+map+=+new+HashMap<String,+Integer>();
for+(String+s+:+list){
++++if+(map.get(s)+==+null){
++++++map.put(s,+1);
++++}+
++++else+{
++++++map.put(s,+map.get(s)+%2B+1);
++++}
}

List<String>+newList+=+new+ArrayList<String>();

//+Remove+from+list+if+there+are+multiples+of+them.
for+(Map.Entry<String,+String>+entry+:+map.entrySet())
{
++if(entry.getValue()+>+1){
++++newList.add(entry.getKey());
++}
}

list.removeAll(newList);|code-block|syntax|javascript|4461432|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Here is a non-Java 8 solution using a map to count occurrences:

<pre><code>Map &lt;String,Integer&gt; map = new HashMap&lt;String, Integer&gt;();
for (String s : list){
 if (map.get(s) == null){
 map.put(s, 1);
 } 
 else {
 map.put(s, map.get(s) + 1);
 }
}

List&lt;String&gt; newList = new ArrayList&lt;String&gt;();

// Remove from list if there are multiples of them.
for (Map.Entry&lt;String, String&gt; entry : map.entrySet())
{
 if(entry.getValue() &gt; 1){
 newList.add(entry.getKey());
 }
}

list.removeAll(newList);
</code></pre>

blocks|key|4461495|text|ArrayList中的解决方案|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4461496|public+static+void+main(String+args[])+throws+Exception+{
++++++List<String>+e+=+new+ArrayList<String>();
++++++List<String>+duplicate+=+new+ArrayList<String>();
++++++e.add("123");
++++++e.add("122");
++++++e.add("125");
++++++e.add("123");

++++++for(String+str+:+e){
++++++++++if(e.indexOf(str)+!=+e.lastIndexOf(str)){
++++++++++++++duplicate.add(str);
++++++++++}
++++++}

++++++for(String+str+:+duplicate){
++++++++++e.remove(str);++++++++++++++
++++++}

++++++for(String+str+:+e){
++++++++++System.out.println(str);
++++++}
++}|code-block|syntax|javascript|4461497|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Solution in ArrayList 

<pre><code>public static void main(String args[]) throws Exception {
 List&lt;String&gt; e = new ArrayList&lt;String&gt;();
 List&lt;String&gt; duplicate = new ArrayList&lt;String&gt;();
 e.add("123");
 e.add("122");
 e.add("125");
 e.add("123");

 for(String str : e){
 if(e.indexOf(str) != e.lastIndexOf(str)){
 duplicate.add(str);
 }
 }

 for(String str : duplicate){
 e.remove(str); 
 }

 for(String str : e){
 System.out.println(str);
 }
 }
</code></pre>

blocks|key|4461503|text|List<String>+e+=+new+ArrayList<String>();
e.add("123");
e.add("122");
e.add("125");
e.add("123");
e.add("125");
e.add("124");
List<String>+sortedList+=+new+ArrayList<String>();
for+(String+current+:+e){
++++if(!sortedList.contains(current)){
++++++++sortedList.add(current);
++++}
++++else{
++++++++sortedList.remove(current);
++++}
}
e.clear();
e.addAll(sortedList);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4461504|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>List&lt;String&gt; e = new ArrayList&lt;String&gt;();
e.add("123");
e.add("122");
e.add("125");
e.add("123");
e.add("125");
e.add("124");
List&lt;String&gt; sortedList = new ArrayList&lt;String&gt;();
for (String current : e){
 if(!sortedList.contains(current)){
 sortedList.add(current);
 }
 else{
 sortedList.remove(current);
 }
}
e.clear();
e.addAll(sortedList);
</code></pre>

blocks|key|4461400|text|使用streams的最简单的解决方案具有O(n%5E2)时间复杂度。如果你在有数百万个条目的List上尝试它们，你会等待很长很长的时间。O(n)解决方案是：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4461401|list+=+list.stream()
+++++++++++.collect(Collectors.groupingBy(Function.identity(),+LinkedHashMap::new,+Collectors.counting()))
+++++++++++.entrySet()
+++++++++++.stream()
+++++++++++.filter(e+->+e.getValue()+==+1)
+++++++++++.map(Map.Entry::getKey)
+++++++++++.collect(Collectors.toList());|code-block|syntax|javascript|4461402|在这里，我使用了一个LinkedHashMap来维护顺序。请注意，静态导入可以简化collect部件。|4461403|这太复杂了，所以我认为使用for循环是解决这个问题的最佳选择。|4461404|Map<String,+Integer>+map+=+new+LinkedHashMap<>();
for+(String+s+:+list)
++++map.merge(s,+1,+Integer::sum);
list+=+new+ArrayList<>();
for+(Map.Entry<String,+Integer>+e+:+map.entrySet())
++++if+(e.getValue()+==+1)
++++++++list.add(e.getKey());|4461405|entityMap^0|K|6|18|4|1U|4|0|0|A|D|15|7|0|D|3|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|10|8|@$9|11|A|12|B|C]|$9|13|A|14|B|C]]|D|@]|E|$]]|$1|M|3|N|5|6|7|15|8|@$9|16|A|17|B|C]]|D|@]|E|$]]|$1|O|3|P|5|H|7|18|8|@]|D|@]|E|$I|J]]|$1|Q|3|-4|5|6|7|19|8|@]|D|@]|E|$]]]|R|$]]

The simplest solutions using streams have <code>O(n^2)</code> time complexity. If you try them on a <code>List</code> with millions of entries, you'll be waiting a very, very long time. An <code>O(n)</code> solution is:

<pre><code>list = list.stream()
 .collect(Collectors.groupingBy(Function.identity(), LinkedHashMap::new, Collectors.counting()))
 .entrySet()
 .stream()
 .filter(e -&gt; e.getValue() == 1)
 .map(Map.Entry::getKey)
 .collect(Collectors.toList());
</code></pre>

Here, I used a <code>LinkedHashMap</code> to maintain the order. Note that static imports can simplify the <code>collect</code> part.

This is so complicated that I think using <code>for</code> loops is the best option for this problem.

<pre><code>Map&lt;String, Integer&gt; map = new LinkedHashMap&lt;&gt;();
for (String s : list)
 map.merge(s, 1, Integer::sum);
list = new ArrayList&lt;&gt;();
for (Map.Entry&lt;String, Integer&gt; e : map.entrySet())
 if (e.getValue() == 1)
 list.add(e.getKey());
</code></pre>

blocks|key|4461382|text|如下所示(使用Set)：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|4461383|Set<Object>+blackList+=+new+Set<>()

public+void+add(Object+object)+{
++++if+(blackList.exists(object))+{
++++++++return;
++++}
++++boolean+notExists+=+set.add(object);
++++if+(!notExists)+{
+++++++set.remove(object)
+++++++blackList.add(object);
++++}
}|code-block|syntax|javascript|4461384|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.oracle.com/javase/7/docs/api/java/util/Set.html^0|7|3|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

Something like this (using a <a href="https://docs.oracle.com/javase/7/docs/api/java/util/Set.html" rel="nofollow">Set</a>):

<pre><code>Set&lt;Object&gt; blackList = new Set&lt;&gt;()

public void add(Object object) {
 if (blackList.exists(object)) {
 return;
 }
 boolean notExists = set.add(object);
 if (!notExists) {
 set.remove(object)
 blackList.add(object);
 }
}
</code></pre>

I have an <code>ArrayList</code> with the following strings;

<pre><code> List&lt;String&gt; e = new ArrayList&lt;String&gt;();
 e.add("123");
 e.add("122");
 e.add("125");
 e.add("123");
</code></pre>

I want to check the list for duplicates and remove them from the list. In this case my list will only have two values, and in this example it would be the values 122 and 125, and the two 123s will go away.

What will be the best way to this? I was thinking of using a <code>Set</code>, but that will only remove one of the duplicates.

Remove duplicates (both values) - duplicate values from an ArrayList

Java

我有一个包含以下字符串的ArrayList； List<String> e = new ArrayList<String>(); e.add("123"); e.add("122"); e.add("125"); e.add("123");我想检查列表中是否有重复项，并将它们从列表中删除。在本例中，我的列表将只有两个值，在本例中，它将是值122和125，而这两个123将消失。最好的方法是什么？我正

问删除重复项(两个值)- ArrayList中的重复值
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回答 7

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