15

谁能告诉我如何删除重复值

String s="Bangalore-Chennai-NewYork-Bangalore-Chennai"; 

和输出应该像

String s="Bangalore-Chennai-NewYork-";

使用Java..

任何帮助,将不胜感激。

4

14 回答 14

38

这在一行中完成:

public String deDup(String s) {
    return new LinkedHashSet<String>(Arrays.asList(s.split("-"))).toString().replaceAll("(^\\[|\\]$)", "").replace(", ", "-");
}

public static void main(String[] args) {
    System.out.println(deDup("Bangalore-Chennai-NewYork-Bangalore-Chennai"));
}

输出:

Bangalore-Chennai-NewYork

请注意,订单被保留:)

要点是:

  • split("-")给我们不同的值作为一个数组
  • Arrays.asList()将数组变成列表
  • LinkedHashSet保留唯一性插入顺序 - 它完成了为我们提供唯一值的所有工作,这些值通过构造函数传递
  • List的toString()[element1, element2, ...]
  • 最后的replace命令从toString()

此解决方案要求值不包含字符序列", "- 对此类简洁代码的合理要求。

Java 8 更新!

当然是1行:

public String deDup(String s) {
    return Arrays.stream(s.split("-")).distinct().collect(Collectors.joining("-"));
}

正则表达式更新!

如果您不关心保留顺序(即可以删除第一次出现的重复项):

public String deDup(String s) {
    return s.replaceAll("(\\b\\w+\\b)-(?=.*\\b\\1\\b)", "");
}
于 2011-07-22T13:35:51.617 回答
4
public static String removeDuplicates(String txt, String splitterRegex)
{
    List<String> values = new ArrayList<String>();
    String[] splitted = txt.split(splitterRegex);
    StringBuilder sb = new StringBuilder();
    for (int i = 0; i < splitted.length; ++i)
    {
        if (!values.contains(splitted[i]))
        {
            values.add(splitted[i]);
            sb.append('-');
            sb.append(splitted[i]);
        }
    }
    return sb.substring(1);

}

用法:

String s = "Bangalore-Chennai-NewYork-Bangalore-Chennai";
s = removeDuplicates(s, "\\-");
System.out.println(s);

印刷:

Bangalore-Chennai-NewYork
于 2011-07-22T13:33:27.037 回答
2

您可以将字符串添加到 HashSet。

  1. 在“-”上拆分字符串。
  2. 将单个单词存储在 Array 中。即 arr[]

小品:

Set<String> set = new HashSet<String>();

    for(int i=0; i < arr.length; i++){
      if(set.contains(arr[i])){
        System.out.println("Duplicate string found at index " + i);
      } else {
        set.add(arr[i]);
      }
于 2011-07-22T13:31:56.307 回答
1

通过拆分创建字符串数组,-然后从中创建一个 hashSet。

String s="Bangalore-Chennai-NewYork-Bangalore-Chennai"; 
String[] strArr = s.split("-");
Set<String> set = new HashSet<String>(Arrays.asList(strArr));

如果要将其作为字符串数组返回,请执行以下操作:

String[] result = new String[set.size()];
set.toArray(result);

这是执行此操作的示例代码:

String s="Bangalore-Chennai-NewYork-Bangalore-Chennai"; 
String[] strArr = s.split("-");
Set<String> set = new LinkedHashSet<String>(Arrays.asList(strArr));
String[] result = new String[set.size()];
set.toArray(result);
StringBuilder res = new StringBuilder();
for (int i = 0; i < result.length; i++) {
    String string = result[i];
    if(i==result.length-1)
        res.append(string);
    else
        res.append(string).append("-");
}
System.out.println(res.toString());

输出:-

Bangalore-Chennai-NewYork
于 2011-07-22T13:34:59.280 回答
1

只是想法:

  1. 解析字符串并使用分隔符“-”拆分标记
  2. 将令牌加载到Collection
  3. 迭代Collection并删除重复项
  4. 使用结果Collection构建新字符串

最棘手的部分应该是 3,但并非不可能。如果您使用 a Set,则可以跳过此步骤。

编辑也许您可以在添加元素之前用存在检查替换 2&3

于 2011-07-22T13:31:53.533 回答
1
static String RemoveDuplicateCharInString(String s){
    for (int i = 0; i < s.length(); i++) {
        if((s.substring(i+1)).indexOf(s.charAt(i))!=-1){
            s=s.substring(0,i+1)+(s.substring(i+1)).replaceAll(""+s.charAt(i),"");
        }
    }
    return s;
}
于 2016-07-01T10:59:00.883 回答
0
wordsArray = s.split("-");
List<String> wordsList = new Arrays.asList(wordsArray);
Set<String>  wordsSet  = new LinkedHashSet<String>(wordsList);

String[] noDuplicates = new String[wordsSet.size()];
wordsSet.toArray(noDuplicates);
于 2011-07-22T13:36:34.080 回答
0

我更喜欢这个比上面所有的都简单。

public void removeDuplicates() {
  String myString = "Bangalore-Chennai-NewYork-Bangalore-Chennai";

  String[] array = myString.split("-");

  Set<String> hashSet = new HashSet<String>(Arrays.asList(array));

  String newString = StringUtils.join(hashSet, "-");        
}
于 2011-07-22T13:42:24.267 回答
0
StringBuilder builderWord = new StringBuilder(word);
for(int index=0; index < builderWord.length(); index++) {
    for(int reverseIndex=builderWord.length()-1; reverseIndex > index;reverseIndex--) {
        if (builderWord.charAt(reverseIndex) == builderWord.charAt(index)) {
            builderWord.deleteCharAt(reverseIndex);
        }
    }
}
return builderWord.toString();
于 2018-07-18T01:29:38.067 回答
0
public class RemDuplicateWordFromString {
public static void main(String[] args) {
    String s1 = "Hello India Hello India Hello India Hello India";
    countWords(s1);
}
public static void countWords(String s1) {
    String[] s2 = s1.split(" ");
    for (int i = 0; i < s2.length; i++) {
        for (int j = i + 1; j < s2.length; j++) {
            if (s2[i].equals(s2[j])) {
                if (i != j) {
                    s2[i] = "";
                }
            }
        }
    }
    for (int i = 0; i < s2.length; i++) {
        if (s2[i] != "") {
            System.out.print(s2[i] + " ");
        }

    }

}

}

于 2017-01-30T11:05:51.410 回答
0
public static void main(String[] args) {
    String str="Bangalore-Chennai-Newyork-Bangalore-Chennai";
    String output="";
    String [] arr=str.split("-");

    LinkedHashSet<String> lhs=new LinkedHashSet<String>();
    for (int i = 0; i < arr.length; i++) {
        lhs.add(arr[i]);
    }
    for(String s:lhs){
        output=output+s+"-";
    }

    System.out.println(output);
}
于 2016-04-23T15:56:44.743 回答
0

游戏有点晚了,但我会简单地使用 HashMap。它很容易理解并且可以快速查找键,可能不是最好的方法,但它仍然是 IMO 的一个很好的答案。当我需要快速和肮脏的格式化时,我一直使用它:

                    String reason = "Word1 , Word2 , Word3";
                    HashMap<String,String> temp_hash = new HashMap<String,String>();
                    StringBuilder reason_fixed = new StringBuilder();
                    //in:
                    for(String word : reason.split(",")){
                        temp_hash.put(word,word);
                    }
                    //out:
                    for(String words_fixed : temp_hash.keySet()){
                        reason_fixed.append(words_fixed + " , ");
                    }
                    //print:
                    System.out.println(reason_fixed.toString());
于 2016-10-12T18:30:31.020 回答
-1
import java.util.HashSet;

public class SplitString {
    public static void main(String[] args) {
        String st = new String("New Delhi-Chennai-New York-Bangalore-Chennai-New Delhi-Chennai-New York");
        StringBuffer stb = new StringBuffer();

        HashSet<String> hashset = new HashSet<String>();
        for (String a : st.split("-"))
            hashset.add(a);

        Object[] str = (Object[]) hashset.toArray();

        for (int i = 0; i < str.length; i++) {
            stb.append(str[i]);

            if (i < str.length - 1)
                stb.append("-");

        }

        System.out.println(stb);
    }
}
于 2014-10-31T13:20:28.310 回答
-2
import java.util.*;

public class RemoveDuplicateWord {

    public static void main(String[] args) {
        String str = "Hai hello Hai how hello are how you";
        removeDupWord(str);
    }

    public static void removeDupWord(String input) {
        List<String> list = Arrays.asList(input.split(" "));
        LinkedHashSet<String> lhs = new LinkedHashSet<String>(list);
        for(String s : lhs) {
            System.out.print(s+" ");
        }                   
    }
}
于 2015-03-31T07:07:09.337 回答