如何从char数组中删除彼此相邻的2个或更多空间

Question

public static void removeDuplicateSpaces(char[] characters) {

    int dupCount = 0;
    for (int i = 0; i < characters.length; i++) {
        if (characters[i] == ' ' && characters[i + 1] == ' ') {
            dupCount++;
            for (int j = i; j < characters.length - 1; j++) {
                characters[j] = characters[j + 1];

            }
            if ((characters[i] == ' ' && characters[i + 1] == ' ')) {
                dupCount++;
                for (int j = i; j < characters.length - 1; j++) {
                    characters[j] = characters[j + 1];
                }
                dupCount++;
            }
            for (int add = characters.length - 1; add > 
                     characters.length- dupCount; add--) {
                characters[add] = '\u0000';

            }

        }

    }

}

I need to reduce all sequences of 2 or more spaces to 1 space within the characters array. If any spaces are removed then the same number of Null character '\' will fill the elements at the end of the array. My code does not remove 4 spaces when there are 5 spaces in between. Such as {'e','','','','','','4'}

Answer 1

How about converting it to String , doing some regex and back to char array?

new String(characters).replaceAll("[ ]+", " ").toCharArray()

EDIT: Ok, I see you're actually not returning the array, but modifying the original array. Then you could use a simple loop

String s = new String(characters).replaceAll("[ ]+", " ");
for (int i = 0; i < characters.length; i++) {
    characters[i] = (i < s.length() ? s.charAt(i) : '\u0000');
}

Answer 2

different variant, to show you how it should work with simple loops:

public static void removeDuplicateSpaces(final char[] characters) {
    for (int i = 0; i < characters.length; i++) {
        while (characters[i] == ' ') { // while current symbol is space
            for (int j = (i + 1); j < characters.length; j++)
                characters[j - 1] = characters[j]; // shift the rest of array
            characters[characters.length - 1] = 0;
        }
    }
}