Question

import java.util.*;
public class DuplicateCharacters {

    public static void main(String[] args) {
        // TODO Auto-generated method stub

        Scanner s = new Scanner(System.in);
        System.out.println("Enter String: ");
        String str = s.nextLine();

        for(int i=0;i<str.length()-1;i++)
        {
            for(int j=i+1;j<str.length();j++)

            {
            if(str.charAt(i)==str.charAt(j))
            {

                System.out.print(str.charAt(j));
            }
            }
        }
    }

}

我的代码适用于字符串仅包含2个重复字符..

e.g。假设输入=印度它打印我但是当输入aaa时，output = a a a ..应该只打印一次

Answer 1

尝试将每个新读取的字符放入Unresolved external symbol _WinMain@16。如果添加到集合返回false，则将其添加到跟踪所有重复字符的集合中。这是伪代码：

Set<Character>

Answer 2

如果可以使用第三方库，则可以使用Eclipse Collections使用以下库：

String str = "india";
MutableCharSet dupes = Strings.asChars(str).toBag().selectDuplicates().toSet();
System.out.println(dupes.makeString(""));  // outputs i

Strings类将从对CharAdapter的调用中返回asChars。 toBag方法将返回一个MutableCharBag，其中包含char中的所有String值及其对应的计数。 selectDuplicates方法将返回另一个MutableCharBag，仅包含出现次数不止一次的char值。 toSet方法返回一个MutableCharSet，它将具有重复的唯一char值。 makeString("")方法将返回存储在MutableCharSet中的字符的字符串，它们之间没有分隔符。

注意：我是Eclipse Collections的提交者。

Answer 3

我首先将问题分解为两部分。首先，您需要一种方法来计算给定起始索引中给定字符的出现次数。这可能看起来像，

static int countFrom(String str, char ch, int start) {
    int count = 0;
    for (int i = start; i < str.length(); i++) {
        if (str.charAt(i) == ch) {
            count++;
        }
    }
    return count;
}

然后您可以将其与Set<Character>一起使用（以跟踪您已使用过的字符），例如

Scanner s = new Scanner(System.in);
System.out.println("Enter String: ");
String str = s.nextLine();
Set<Character> set = new HashSet<>();
for (int i = 0; i < str.length(); i++) {
    char ch = str.charAt(i);
    if (!set.contains(ch)) {
        int count = countFrom(str, ch, i);
        if (count > 1) {
            System.out.print(ch);
        }
        set.add(ch);
    }
}
System.out.println();

如果没有Set，您可以模仿String之类的

String setString = "";
for (int i = 0; i < str.length(); i++) {
    char ch = str.charAt(i);
    if (!setString.contains(String.valueOf(ch))) {
        int count = countFrom(str, ch, i);
        if (count > 1) {
            System.out.print(ch);
        }
        setString += ch;
    }
}

Answer 4

这是解决它的另一种方法..

    Scanner s = new Scanner(System.in);
    System.out.println("Enter String: ");
    String str = s.nextLine();

    String duplicates = "";

    for (int i = 0; i < str.length() - 1; i++) {
        for (int j = i + 1; j < str.length(); j++)

        {
            if (str.charAt(i) == str.charAt(j)) {

                if (!duplicates.contains(String.valueOf(str.charAt(j)))) {
                    duplicates += str.charAt(j);
                    break;
                }

            }
        }
    }

    System.out.println(duplicates);

Answer 5

或考虑一张地图

    String word = "aiaiiab";
    Map <Character, Integer> charMap = new HashMap<Character, Integer> ();
    for (char c : word.toCharArray()) {
        Integer count = charMap.get(new Character(c));
        if (count == null) {
            charMap.put(new Character(c), 1);
        }
        else {
            charMap.put(new Character(c), count.intValue() + 1);
        }
    }

    for (Entry<Character, Integer> obj : charMap.entrySet()) {

        if (obj.getValue() > 1) {
            System.out.println(obj.getKey());
        }
    }

Answer 6

package com.java2novice.algos;

import java.util.HashMap;
import java.util.Map;
import java.util.Set;

public class DuplicateCharsInString {

public void findDuplicateChars(String str){

    Map<Character, Integer> dupMap = new HashMap<Character, Integer>();
    char[] chrs = str.toCharArray();
    for(Character ch:chrs){
        if(dupMap.containsKey(ch)){
            dupMap.put(ch, dupMap.get(ch)+1);
        } else {
            dupMap.put(ch, 1);
        }
    }
    Set<Character> keys = dupMap.keySet();
    for(Character ch:keys){
        if(dupMap.get(ch) > 1){
            System.out.println(ch+"--->"+dupMap.get(ch));
        }
    }
}

public static void main(String a[]){
    DuplicateCharsInString dcs = new DuplicateCharsInString();
    dcs.findDuplicateChars("Java2Novice");
}
}

Answer 7

public class GFG  
{ 
  static final int NO_OF_CHARS = 256; 

  /* Fills count array with frequency of characters */
  static void fillCharCounts(String str, int[] count) 
  { 
    for (int i = 0; i < str.length();  i++) 
      count[str.charAt(i)]++; 
  } 

  /* Print duplicates present in the passed string */
  static void printDups(String str) 
  { 
    // Create an array of size 256 and fill count of every character in it 
    int count[] = new int[NO_OF_CHARS]; 
    fillCharCounts(str, count); 

    for (int i = 0; i < NO_OF_CHARS; i++) 
      if(count[i] > 1) 
        System.out.printf("%c,  count = %d \n", i,  count[i]); 
  } 

  // Driver Method 
  public static void main(String[] args) 
  { 
    String str = "test string"; 
    printDups(str); 
  } 
}

Answer 8

import java.util.*;

public class DuplicateCharacters {

    static final int NO_OF_CHARS = 256;

    public static void main(String[] args) {

       Scanner s = new Scanner(System.in);
       System.out.println("Enter String: ");
       String str = s.nextLine();
       int count[] = new int[NO_OF_CHARS];

       for(int i=0;i<str.length();i++)
       {
          count[str.charAt(i)]++;
       }
       for (int i = 0; i < NO_OF_CHARS; i++) 
          if(count[i] > 1) 
             System.out.printf("%c", i);
      }
   }

Answer 9

转换char数组中的String并将其附加到String builder中，并检查String builder是否已包含该char，然后打印该重复的char。

public static void findduplicate(String word) {

        char arr[] = word.toCharArray();
        StringBuilder s = new StringBuilder();
        for (char c : arr) {
            int index = s.indexOf("" + c);
            if (index != -1) {
                System.out.println("duplicate character "+c);
            } else {
                s.append(c);
            }
        }

    }

如何在java

9 个答案: