在Java中查找具有连续字符的字符串 [英] Finding strings with consecutive characters in Java

问题描述

在Java中编写一个函数，它接受一个字符串数组，并从字符串数组中返回那些连续重复特定字母的字符串，例如：如果I / P是

  {Dauresselam，slab，fuss，boolean，clap} 
  

那么O / P应该是

  {Dauresselam， fuss，boolean} 
  

我可以使用

解决它

  import java.util.Scanner; 
 public class doubleChars {
 public static String [] getDoubles（String [] In）
 {
 
 int inLen = In.length; 
 String zoom [] = new String [inLen]; 
 int count = 0; 
 if（inLen == 0）
 {
 return zoom; 
} 
 for（int i = 0; i< = inLen-1; i ++）
 {
 String A = In [i]; 
 //System.out.println(A); 
 int striLen = A.length（）; 
 for（int j = 0; j< striLen-1; j ++）
 {
 
 if（A.substring（j，j + 1）.equals（A.substring （j + 1，j + 2）））
 {
 zoom [count] = A; 
 count ++; 
休息; 
} 
} 
 
} 
返回缩放; 
} 
 public static void main（String [] args）
 {
 char more ='y'; 
 int ab = 0; 
 String [] res = {}; 
 String [] fillMe = {durres，murres，，abcdeee，boolean，nger，lagger}; 
 Scanner strobe = new Scanner（System.in）; 
 System.out.println（请输入字符串的数组）; 
 / * while（strobe.hasNext（））
 {
 fillMe [ab] = strobe.next（）; 
 
 ab ++; 
} 
 * / 
 res = doubleChars.getDoubles（fillMe）; 
 for（int k = 0; k< res.length; k ++）
 {
 if（res [k] == null）
 {
 break; 
} 
 System.out.println（res [k]）; 
} 
 
 
} 
} 
  

IS 有没有办法使用正则表达式来缩短它？

解决方案

你可以用一个

---

Java示例：

  String [] strings = {Dauresselam，slab ，fuss，boolean，clap}; 
 
 String regex =（[a-z]）\\1; 
模式模式= Pattern.compile（正则表达式）; 
 
 for（String string：strings）{
 Matcher matcher = pattern.matcher（string）; 
 if（matcher.find（））{
 System.out.println（string）; 
} 
} 
  

打印：

  Dauresselam 
 fuss 
 boolean 
  

Write a function in Java which takes an Array of strings and from the array of strings returns only those strings which have a consecutive repetition of a particular letter for eg: if I/P is

{"Dauresselam", "slab", "fuss", "boolean", "clap"}

then O/P should be

{"Dauresselam", "fuss", "boolean"}

I could solve it using

import java.util.Scanner;
public class doubleChars {
public static String[] getDoubles(String[]In)
{

    int inLen=In.length;
    String zoom[]=new String[inLen];
    int count=0;
    if(inLen==0)
    {
        return zoom;
    }
    for(int i=0;i<=inLen-1;i++)
    {
        String A=In[i];
        //System.out.println(A);
        int striLen=A.length();
        for(int j=0;j<striLen-1;j++)
        {

            if(A.substring(j, j+1).equals(A.substring(j+1, j+2)))
            {
                zoom[count]=A;
                count++;
                break;
            }
        }

    }
      return zoom;
    }
 public static void main(String[] args)
 {
     char more='y';
     int ab=0;
    String[] res={};
    String[] fillMe={"durres", "murres", "", "abcdeee", "boolean", "nger", "lagger"};
    Scanner strobe=new Scanner(System.in);
    System.out.println("Please enter the arraye of the string");
    /*while(strobe.hasNext())
    {
        fillMe[ab]=strobe.next();

        ab++;
    }
    */
    res=doubleChars.getDoubles(fillMe);
    for(int k=0;k<res.length;k++)
    {
        if(res[k]==null)
        {
            break;
        }
    System.out.println(res[k]);
    }


}
}

IS there a way to use regex to make it shorter?

解决方案

You could use a backreference:

([a-z])\1

Visualization by Debuggex

---

Java example:

String[] strings = { "Dauresselam", "slab", "fuss", "boolean", "clap" };

String regex = "([a-z])\\1";
Pattern pattern = Pattern.compile(regex);

for (String string : strings) {
    Matcher matcher = pattern.matcher(string);
    if (matcher.find()) {
        System.out.println(string);
    }
}

Prints:

Dauresselam
fuss
boolean

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