blocks|key|4490670|text|您可以对字符串的字符数组(str.toCharArray+())进行排序，并从排序后的数组创建一个新字符串，以获得字符串的“规范”表示。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4490671|然后，您可以将这些字符串添加到Set中，并检查每个字符串的规范表示是否已经存在于集合中。|4490672|Set<String>+letter+=+new+HashSet<String>();
for+(int+i+=+0;+i+<+name1.length;+i%2B%2B)+{
++++char[]+chars+=+name1[i].toCharArray();
++++Arrays.sort(chars);
++++letter.add(new+String(chars));
}
for+(int+j+=+0;+j+<+name2.length;+j%2B%2B)+{
++++char[]+chars+=+name2[j].toCharArray();
++++Arrays.sort(chars);
++++letter.add(new+String(chars));
}
for+(int+k+=+0;+k+<+name3.length;+k%2B%2B)+{
++++char[]+chars+=+name3[k].toCharArray();
++++Arrays.sort(chars);
++++letter.add(new+String(chars));
}|code-block|syntax|javascript|4490673|编辑:我将Set<char[]>更改为Set<String>，因为数组不覆盖hashCode和equals，所以HashSet<char[]>不起作用。|4490674|entityMap^0|D|I|0|F|3|0|0|5|B|J|B|12|8|1B|6|1K|F|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|T|8|@$9|U|A|V|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|W|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|X|8|@$9|Y|A|Z|B|C]|$9|10|A|11|B|C]|$9|12|A|13|B|C]|$9|14|A|15|B|C]|$9|16|A|17|B|C]]|D|@]|E|$]]|$1|O|3|-4|5|6|7|18|8|@]|D|@]|E|$]]]|P|$]]

You can sort the character array of the String (<code>str.toCharArray ()</code>) and create a new String from the sorted array to get a "canonical" representation of a String.

Then you can add these Strings to a <code>Set</code>, and check for each String whether the canonical representation is already in the Set.

<pre><code>Set&lt;String&gt; letter = new HashSet&lt;String&gt;();
for (int i = 0; i &lt; name1.length; i++) {
 char[] chars = name1[i].toCharArray();
 Arrays.sort(chars);
 letter.add(new String(chars));
}
for (int j = 0; j &lt; name2.length; j++) {
 char[] chars = name2[j].toCharArray();
 Arrays.sort(chars);
 letter.add(new String(chars));
}
for (int k = 0; k &lt; name3.length; k++) {
 char[] chars = name3[k].toCharArray();
 Arrays.sort(chars);
 letter.add(new String(chars));
}
</code></pre>

EDIT : I changed the <code>Set&lt;char[]&gt;</code> to <code>Set&lt;String&gt;</code>, since arrays don't override <code>hashCode</code> and <code>equals</code>, so <code>HashSet&lt;char[]&gt;</code> wouldn't work.

blocks|key|4491665|text|TreeSet允许我们提供一个比较器。看看这是否有帮助。为了保持计数，请使用TreeMap。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4491666|package+empty;

import+java.util.Arrays;
import+java.util.Comparator;
import+java.util.Set;
import+java.util.TreeMap;
import+java.util.TreeSet;

public+class+RemoveDuplicateStrings+{

++++public+static+void+main(String[]+args)+{
++++++++String[]+name1+=+{+"amy",+"jose",+"jeremy",+"alice",+"patrick"+};
++++++++String[]+name2+=+{+"alan",+"may",+"jeremy",+"helen",+"alexi"+};
++++++++String[]+name3+=+{+"adel",+"aron",+"amy",+"james",+"yam"+};

++++++++Comparator<String>+comparator+=+new+Comparator<String>()+{
++++++++++++@Override+public+int+compare(String+o1,+String+o2)+{
++++++++++++++++System.out.println("Compare("+%2B+o1+%2B+","+%2B+o2+%2B+")");
++++++++++++++++char[]+a1+=+o1.toCharArray();
++++++++++++++++Arrays.sort(a1);
++++++++++++++++char[]+a2+=+o2.toCharArray();
++++++++++++++++Arrays.sort(a2);
++++++++++++++++return+new+String(a1).compareTo(new+String(a2));
++++++++++++}
++++++++};
++++++++Set<String>+set+=+new+TreeSet<String>(comparator);

++++++++for+(String+name+:+name1)+{
++++++++++++set.add(name);
++++++++}
++++++++for+(String+name+:+name2)+{
++++++++++++set.add(name);
++++++++}
++++++++for+(String+name+:+name3)+{
++++++++++++set.add(name);
++++++++}

++++++++String[]+result+=+set.toArray(new+String[set.size()]);
++++++++System.out.println(Arrays.asList(result));

++++++++//+Using+TreeMap+to+keep+the+count.

++++++++TreeMap<String,+Integer>+map+=+new+TreeMap<String,+Integer>(comparator);

++++++++addAll(name1,+map);
++++++++addAll(name2,+map);
++++++++addAll(name3,+map);

++++++++System.out.println(map);
++++}

++++private+static+void+addAll(String[]+names,+TreeMap<String,+Integer>+map)+{
++++++++for+(String+name+:+names)+{
++++++++++++if+(map.containsKey(name))+{
++++++++++++++++int+n+=+map.get(name);
++++++++++++++++map.put(name,+n+%2B+1);
++++++++++++}+else
++++++++++++++++map.put(name,+1);
++++++++}
++++}
}|code-block|syntax|javascript|4491667|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.oracle.com/javase/7/docs/api/java/util/TreeSet.html|1|https://docs.oracle.com/javase/7/docs/api/java/util/TreeMap.html^0|0|7|12|7|0|7|0|12|7|1|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@$9|Z|A|10|1|11]|$9|12|A|13|1|14]]|E|$]]|$1|F|3|G|5|H|7|15|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|16|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]|S|$5|N|O|P|E|$Q|T]]]]

<a href="https://docs.oracle.com/javase/7/docs/api/java/util/TreeSet.html" rel="nofollow"><code>TreeSet</code></a> allows us to give a comparator. See whether this helps. For keeping the count use a <a href="https://docs.oracle.com/javase/7/docs/api/java/util/TreeMap.html" rel="nofollow"><code>TreeMap</code></a>.

<pre><code>package empty;

import java.util.Arrays;
import java.util.Comparator;
import java.util.Set;
import java.util.TreeMap;
import java.util.TreeSet;

public class RemoveDuplicateStrings {

 public static void main(String[] args) {
 String[] name1 = { "amy", "jose", "jeremy", "alice", "patrick" };
 String[] name2 = { "alan", "may", "jeremy", "helen", "alexi" };
 String[] name3 = { "adel", "aron", "amy", "james", "yam" };

 Comparator&lt;String&gt; comparator = new Comparator&lt;String&gt;() {
 @Override public int compare(String o1, String o2) {
 System.out.println("Compare(" + o1 + "," + o2 + ")");
 char[] a1 = o1.toCharArray();
 Arrays.sort(a1);
 char[] a2 = o2.toCharArray();
 Arrays.sort(a2);
 return new String(a1).compareTo(new String(a2));
 }
 };
 Set&lt;String&gt; set = new TreeSet&lt;String&gt;(comparator);

 for (String name : name1) {
 set.add(name);
 }
 for (String name : name2) {
 set.add(name);
 }
 for (String name : name3) {
 set.add(name);
 }

 String[] result = set.toArray(new String[set.size()]);
 System.out.println(Arrays.asList(result));

 // Using TreeMap to keep the count.

 TreeMap&lt;String, Integer&gt; map = new TreeMap&lt;String, Integer&gt;(comparator);

 addAll(name1, map);
 addAll(name2, map);
 addAll(name3, map);

 System.out.println(map);
 }

 private static void addAll(String[] names, TreeMap&lt;String, Integer&gt; map) {
 for (String name : names) {
 if (map.containsKey(name)) {
 int n = map.get(name);
 map.put(name, n + 1);
 } else
 map.put(name, 1);
 }
 }
}
</code></pre>

blocks|key|4490720|text|与kdm一致：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4490721|import+java.util.ArrayList;
import+java.util.Arrays;
import+java.util.HashSet;
import+java.util.List;
import+java.util.Set;

public+class+RemoveDuplicateString+{

++++private+static+boolean+add(Set<String>+keySet,+String+s){
++++++++char[]+sortCharacters+=+s.toCharArray();
++++++++Arrays.sort(sortCharacters);
++++++++return+keySet.add(new+String(sortCharacters));
++++}

++++private+static+void+check(Set<String>+keySet,+String+[]names,+List<String>+result){
++++++++for+(String+name+:+names)+{
++++++++++++if+(add(keySet,+name)){
++++++++++++++++result.add(name);
++++++++++++}
++++++++}
++++}

++++public+static+void+main(String[]+args)+{
++++++++String[]+name1+=+{"amy",+"jose",+"jeremy",+"alice",+"patrick"};
++++++++String[]+name2+=+{"alan",+"may",+"jeremy",+"helen",+"alexi"};
++++++++String[]+name3+=+{"adel",+"aron",+"amy",+"james",+"yam"};
++++++++Set<String>+keySet+=+new+HashSet<String>();
++++++++List<String>+result+=+new+ArrayList<String>();
++++++++check(keySet,+name1,+result);
++++++++check(keySet,+name2,+result);
++++++++check(keySet,+name3,+result);
++++++++System.out.println(result);
++++}
}|code-block|syntax|javascript|4490722|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|$]]

In line with kdm:

<pre><code>import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashSet;
import java.util.List;
import java.util.Set;

public class RemoveDuplicateString {

 private static boolean add(Set&lt;String&gt; keySet, String s){
 char[] sortCharacters = s.toCharArray();
 Arrays.sort(sortCharacters);
 return keySet.add(new String(sortCharacters));
 }

 private static void check(Set&lt;String&gt; keySet, String []names, List&lt;String&gt; result){
 for (String name : names) {
 if (add(keySet, name)){
 result.add(name);
 }
 }
 }

 public static void main(String[] args) {
 String[] name1 = {"amy", "jose", "jeremy", "alice", "patrick"};
 String[] name2 = {"alan", "may", "jeremy", "helen", "alexi"};
 String[] name3 = {"adel", "aron", "amy", "james", "yam"};
 Set&lt;String&gt; keySet = new HashSet&lt;String&gt;();
 List&lt;String&gt; result = new ArrayList&lt;String&gt;();
 check(keySet, name1, result);
 check(keySet, name2, result);
 check(keySet, name3, result);
 System.out.println(result);
 }
}
</code></pre>

I'm facing a problem right now. In one of my program, I need to remove strings with same characters from an Array. For eg. suppose,

I have 3 Arrays like,

<pre><code>String[] name1 = {"amy", "jose", "jeremy", "alice", "patrick"};
String[] name2 = {"alan", "may", "jeremy", "helen", "alexi"};
String[] name3 = {"adel", "aron", "amy", "james", "yam"};
</code></pre>

As you can see, there is a String <code>amy</code> in the <code>name1</code> array. Also, I have strings like <code>may</code>, <code>amy</code> and <code>yam</code> in the next two arrays. What I need is that, I need a final array which does not contain these repeated Strings. I just need only one occurrence: I need to remove all permutations of a name in the final array. That is the final array should be:

<pre><code>String[] finalArray={"amy", "jose", "alice", "patrick","alan", "jeremy", "helen", "alexi","adel", "aron", "james"}
</code></pre>

(The above array removed both yam, may, and only includes amy). 

What i have tried so far, using <code>HashSet</code>, is as below

<pre><code>String[] name1 = {"Amy", "Jose", "Jeremy", "Alice", "Patrick"};
String[] name2 = {"Alan", "mAy", "Jeremy", "Helen", "Alexi"};
String[] name3 = {"Adel", "Aaron", "Amy", "James", "Alice"};
Set&lt;String&gt; letter = new HashSet&lt;String&gt;();
for (int i = 0; i &lt; name1.length; i++) {
 letter.add(name1[i]);
}
for (int j = 0; j &lt; name2.length; j++) {
 letter.add(name2[j]);
}
for (int k = 0; k &lt; name3.length; k++) {
 letter.add(name3[k]);
}
System.out.println(letter.size() + " letters must be sent to: " + letter);
</code></pre>

But, the problem with this code is that, it just removes multiple occurences of the same string. Is there any other alternative? Any help is very much appreciated.

Remove Strings with same characters in a String Array

我现在正面临着一个问题。在我的一个程序中，我需要从数组中删除具有相同字符的字符串。例如。假设，我有3个数组，例如，String[] name1 = {"amy", "jose", "jeremy", "alice", "patrick"};String[] name2 = {"alan", "may", "jeremy...

问删除字符串数组中具有相同字符的字符串
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