How can I split a string into words in Java without using String.split()?

Question

My teacher specifically requested that we split a sentence into words without using String.split() . I've done it using a Vector (which we haven't learned), a while -loop, and substrings. What are other ways of accomplishing this? (preferably without using Vectors / ArrayLists ).

Answer 1

I believe that your teacher is asking you to process the string yourself (without using any other libraries to do it for you). Check to see if this is the case - if you can use them, there are things such as StringTokenizer, Pattern, and Scanner to facilitate string processing.

Otherwise...

You will need a list of word separators (such as space, tab, period, etc...) and then walk the array, building a string a character at a time until you hit the word separator. After finding a complete word (you have encountered a word separator character), save it the variable out into your structure (or whatever is required), reset the variable you are building the word in and continue.

Answer 2

Parsing the string character by character, copying each character into a new String, and stopping when you reach a white space character. Then start a new string and continue until you reach the end of the original string.

Answer 3

您可以使用java.util.Scanner。

Answer 4

You can use java.util.StringTokenizer to split a text using desired delimiter. Default delimiter is SPACE/TAB/NEW_LINE.

String myTextToBeSplit = "This is the text to be split into words.";  
StringTokenizer tokenizer = new StringTokenizer( myTextToBeSplit );  
while ( tokinizer.hasMoreTokens()) {  
    String word = tokinizer.nextToken();  
    System.out.println( word ); // word you are looking in  
}  

As an alternate you can also use java.util.Scanner

Scanner s = new Scanner(myTextToBeSplit).useDelimiter("\\s");  
while( s.hasNext() ) {  
System.out.println(s.next());  
}  
s.close();  

Answer 5

import java.util.Arrays;
public class ReverseTheWords {

    public static void main(String[] args) {
        String s = "hello java how do you do";
        System.out.println(Arrays.toString(ReverseTheWords.split(s)));
    }

    public static String[] split(String s) {
        int count = 0;
        char[] c = s.toCharArray();

        for (int i = 0; i < c.length; i++) {
            if (c[i] == ' ') {
                count++;
            }
        }
        String temp = "";
        int k = 0;
        String[] rev = new String[count + 1];
        for (int i = 0; i < c.length; i++) {
            if (c[i] == ' ') {
                rev[k++] = temp;
                temp = "";
            } else
                temp = temp + c[i];
        }
        rev[k] = temp;
        return rev;
    }

}

Answer 6

你可以使用StringTokenizer http://www.java-samples.com/showtutorial.php?tutorialid=236

Answer 7

或者使用Pattern （也称为正则表达式）来尝试匹配单词。

Answer 8

• Use a Scanner with ctor (String)
• regular expressions and match
• StringTokenizer
• iterating yourself char by char
• recursive iteration

Answer 9

Without using a Vector / List (and without manually re-implementing their ability to re-size themselves for your function), you can take advantage of the simple observation that a string of length N cannot have more than (N+1)/2 words (in integer division). You can declare an array of strings of that size, populate it the same way you populated that Vector , and then copy the results to an array of the size of the number of words you found.

So:

String[] mySplit( String in ){
    String[] bigArray = new String[ (in.length()+1)/2 ];

    int numWords = 0;
    // Populate bigArray with your while loop and keep
    // track of the number of words

    String[] result = new String[numWords];
    // Copy results from bigArray to result

    return result;
}

Answer 10

public class MySplit {

public static String[] mySplit(String text,String delemeter){
    java.util.List<String> parts = new java.util.ArrayList<String>();
    text+=delemeter;        

    for (int i = text.indexOf(delemeter), j=0; i != -1;) {
        parts.add(text.substring(j,i));
        j=i+delemeter.length();
        i = text.indexOf(delemeter,j);
    }


    return parts.toArray(new String[0]);
}

public static void main(String[] args) {
    String str="012ab567ab0123ab";
    String delemeter="ab";
    String result[]=mySplit(str,delemeter);
    for(String s:result)
        System.out.println(s);
}

}

Answer 11

public class sha1 {
public static void main(String[] args) {
    String s = "hello java how do you do";
    System.out.println(Arrays.toString(sha1.split(s)));
}
public static String[] split(String s) {
    int count = 0;
    char[] c = s.toCharArray();

    for (int i = 0; i < c.length; i++) {
        if (c[i] == ' ') {
            count++;
        }
    }
    String temp = "";
    int k = 0;
    String[] rev = new String[count + 1];
    for (int i = c.length-1; i >= 0; i--) {
        if (c[i] == ' ') {
            rev[k++] = temp;
            temp = "";
        } else
            temp = temp + c[i];
    }
    rev[k] = temp;
    return rev;
}

}

Answer 12

Simple touch.! Improve if you want to.

package com.asif.test;

public class SplitWithoutSplitMethod {

public static void main(String[] args) {

    split('@',"asif@is@handsome");
}

static void split(char delimeter, String line){

    String word = "";
    String wordsArr[] = new String[3];

    int k = 0;
    for(int i = 0; i <line.length(); i++){

        if(line.charAt(i) != delimeter){
            word+= line.charAt(i);
        }else{
            wordsArr[k] = word;
            word = "";
            k++;                    
        }
    }
    wordsArr[k] = word;
    for(int j = 0; j <wordsArr.length; j++)
    System.out.println(wordsArr[j]);

}

}

Answer 13

Please try this .

  public static String[] mysplit(String mystring) {

    String string=mystring+" ";               //append " " bcz java string does not hava any ending character
    int[] spacetracker=new int[string.length()];// to count no. of spaces in string
    char[] array=new char[string.length()];     //store all non space character
    String[] tokenArray=new String[string.length()];//to return token of words

    int spaceIndex=0;
    int parseIndex=0;
    int arrayIndex=0;
    int k=0;
    while(parseIndex<string.length())
    {
        if(string.charAt(parseIndex)==' '||string.charAt(parseIndex)==' ')
        {
            spacetracker[spaceIndex]=parseIndex;
            spaceIndex++;
            parseIndex++;
        }else
        {
        array[arrayIndex]=string.charAt(parseIndex);
        arrayIndex++;
        parseIndex++;
        }
    }


    for(int i=0;i<spacetracker.length;i++)
    {
        String token="";
        for(int j=k;j<(spacetracker[i])-i;j++)
        {
            token=token+array[j];
            k++;

        }
        tokenArray[i]=token;
        //System.out.println(token);
        token="";

    }
    return tokenArray;
}

Hope this helps

Answer 14

import java.util.*;
class StringSplit {
    public static void main(String[] args) 
    {
        String s="splitting a string without using split()";
        ArrayList<Integer> al=new ArrayList<Integer>();     //Instead you can also use a String 
        ArrayList<String> splitResult=new ArrayList<String>();
        for(int i=0;i<s.length();i++)
            if(s.charAt(i)==' ')
                al.add(i);
        al.add(0, 0);
        al.add(al.size(),s.length());
        String[] words=new String[al.size()];
        for(int j=0;j<=words.length-2;j++)
                splitResult.add(s.substring(al.get(j),al.get(j+1)).trim());
        System.out.println(splitResult);
    }
}

Time complexity: O(n)

Answer 15

You can use java Pattern to do it in easy way.

package com.company;

import java.util.regex.Pattern;

public class umeshtest {

    public static void main(String a[]) {
        String ss = "I'm Testing and testing the new feature";
        Pattern.compile(" ").splitAsStream(ss).forEach(s -> System.out.println(s));
    }
}

Answer 16

您还可以使用String.substring或charAt[] 。