如何比较单个字符串中的元素？

Question

I need to print the indexes of the elements in a string of 1s and 0s if and only if the 1s and 0s repeat more than once side by side.

For example: input: "0010011" output: "0, 3, 5"

A repeat starts at indexes 0 , 3, and 5.

Here is what I currently have

Scanner keyboard = new Scanner(System.in);
    ArrayList runs = new ArrayList();


    System.out.println("Enter non empty string of 1s and 0s");
    String input = keyboard.nextLine();
    char[] array = input.toCharArray();

    for(int i = 0; i < input.length(); i++)
    {
        if(array[i] == array[i++])
        {
            runs.add(i);
        }

    }
    for(int i = 0; i<= array.length; i++)
    {
        System.out.println(runs);
    }

When I test this, I try input of "00100" and get an output of "1,3,5". Another test input from above "0010011" and get "1,3,5,7". It seems to be printing odd numbers, not the indexes where numbers start to repeat. Can anyone spot what I am doing wrong? I have a feeling its coming from my comparing in the first for loop.

Answer 1

This should work:

Scanner keyboard = new Scanner(System.in);

System.out.println("Enter non empty string of 1s and 0s");
String input = keyboard.nextLine();
char lastChar = '~'

for(int i=0; i<input.length() - 1; i++) {
    if(input.charAt(i) == input.charAt(i + 1) && (i == 0 || input.charAt(i) != input.charAt(i - 1))) {
        System.out.println(i);
    }
}

Answer 2

There are 3 immediate issues that I can point out:

• First is obvious. You are using i++ , while you should be using i + 1 . i++ will increment the value of i , so you're missing out an index there for next iteration.
• Second, you should iterate till input.length() - 2 , else you will get an ArrayIndexOutOfBounds exception, when you access array[i + 1] for the last index.
• Your current logic will fail if you've more than 2 0's or 1's in sequence. It will print both 0 and 1 index for say, 00010 .

For the third point, what you should do is, once you find two consecutive characters same, you should skip upcoming character that are same. You will need an inner loop here. Probably a do-while .

You should modify your for loop to this:

for(int i = 0; i < input.length() - 1; i++)
{
    if(array[i] == array[i+1])
    {
        System.out.println(i);
        do {
            i++;
        } while (i < input.length() - 1 && array[i] == array[i + 1]);
    }
}

Also, ArrayList doesn't have a length attribute. You should use size() method to get maximum size.

---

BTW, this can also be done using regex. Well, you might not have been taught about this yet, but this is just for another possible way:

Pattern pattern = Pattern.compile("0{2,}|1{2,}");
Matcher matcher = pattern.matcher(input);

while (matcher.find()) {
    System.out.println(matcher.start());
}

Answer 3

Two problems there:

First, when the postfix ++ return the value before take effect, so you are comparing array[i] with itself.

Also, you are doing two plus 1 each loop, so there are only odd numbers.

Answer 4

Try this :-

Scanner keyboard = new Scanner(System.in);
    ArrayList runs = new ArrayList();


    System.out.println("Enter non empty string of 1s and 0s");
    String input = keyboard.nextLine();
    char[] array = input.toCharArray();

    for(int i = 0; i < input.length()-1; i++)
    {
        if(array[i] == array[i+1])
        {
            runs.add(i);
        }

    }
    for(int i = 0; i<= array.length; i++)
    {
        System.out.println(runs);
    }