Finds the index of the first word that starts with the given prefix. - Android java.lang

Android examples for java.lang:String Search

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Description

Finds the index of the first word that starts with the given prefix.

Demo Code

import android.text.TextUtils;
import java.io.UnsupportedEncodingException;
import java.security.MessageDigest;
import java.security.NoSuchAlgorithmException;
import java.util.ArrayList;
import java.util.Collection;
import java.util.Iterator;
import java.util.List;
import java.util.Map;
import java.util.Random;
import java.util.UUID;
import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class Main{

    /**/*from w  w w .ja v a2  s .c om*/
     * Finds the index of the first word that starts with the given prefix.
     * <p>
     * 
     * @param text
     *            the text in which to search for the prefix
     * @param prefix
     *            the text to find, in upper case letters
     * @return index If not found,returns -1.
     */
    public static int indexOfWordPrefix(CharSequence text, char[] prefix) {
        if (prefix == null || text == null) {
            return -1;
        }

        int textLength = text.length();
        int prefixLength = prefix.length;

        if (prefixLength == 0 || textLength < prefixLength) {
            return -1;
        }

        int i = 0;
        while (i < textLength) {
            // Skip non-word characters
            while (i < textLength
                    && !Character.isLetterOrDigit(text.charAt(i))) {
                i++;
            }

            if (i + prefixLength > textLength) {
                return -1;
            }

            // Compare the prefixes
            int j;
            for (j = 0; j < prefixLength; j++) {
                if (Character.toUpperCase(text.charAt(i + j)) != prefix[j]) {
                    break;
                }
            }
            if (j == prefixLength) {
                return i;
            }

            // Skip this word
            while (i < textLength
                    && Character.isLetterOrDigit(text.charAt(i))) {
                i++;
            }
        }

        return -1;
    }

}

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