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4 digits numbers in an array as String


4 digits numbers in an array as String

By : Kasaragadda Srinu
Date : November 22 2020, 10:38 AM
I think the issue was by ths following , I need a String array with the following attributes: , The following code will generate an array with your specifications.
code :
public class Test {

public static void main(String[] args) {

    List<String> result = new ArrayList<>();

    Set<Character> set = new HashSet<>();
    for (int i = 1234; i <= 9876; i++) {
        set.clear();

        String iAsString = Integer.toString(i);
        char[] chars = iAsString.toCharArray();

        boolean valid = true;
        for (char c : chars) {
            if (c == '0' || !set.add(c)) {
                valid = false;
                break;
            }
        }

        if (valid) {
            result.add(iAsString);
        }
    }


    String[] yourStringArray = result.toArray(new String[result.size()]);

    System.out.println(Arrays.toString(yourStringArray));

}


}


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Convert digits from string of numbers into ints then into array

Convert digits from string of numbers into ints then into array


By : vjD
Date : March 29 2020, 07:55 AM
With these it helps I have a 1000 digit number that I am storing as a string. I want to end up with each digit in that number as its own element in a vector. so "12345" would be {1,2,3,4,5}. Here is the code I have so far, but it is not working. The error is coming from comparing the string letter to the number. Any help would be appreciated. I have seen another question that is similar, but the string contains spaces between numbers, this doesn't. , To compare char in string use
code :
input_string[i] == '0'
if (&input_string[i] == "0")
if (input_string[i] == '0')  //Removed & from &input_string[i], and replace double quotation(") to single quotation(').
return_vector.reserve(input_string.length());  //Requests  vector capacity be at least enough to contain input_string.length() number of element, which reduce redundant reallocation.

for (int i = 0; i<input_string.length(); i++)
{
    return_vector.push_back(input_string[i]-'0');
}
Split string of numbers into digits and send to array

Split string of numbers into digits and send to array


By : Ishaq Syed
Date : March 29 2020, 07:55 AM
With these it helps I want to split a 40 digit number and send each digit into an array called digits. ToArray doesn't work because it can't convert from char to int. Neither does Split I think? I'm stumped
code :
string digits = "8957853759876839473454789595495735984339";
int[] array = digits.Select(x => (int)char.GetNumericValue(x)).ToArray();
int[] array = digits.Select(x => x - 48).ToArray();
int[] array = digits.Where(x => char.IsNumber(x)).Select(x => x - 48).ToArray();
Saving numbers (sequence of digits) from a string in an array of pointers to a char

Saving numbers (sequence of digits) from a string in an array of pointers to a char


By : Sherbban
Date : March 29 2020, 07:55 AM
This might help you I am supposed to save every sequence of digits from a string in an array of chars , this is what i tried: , Some issues:
code :
ar[index++]=temp;
char temp[100] = {0}
temp[i] = '\0';
char * ar[10];
void int_in_string(char *s, char **ar, int MaxCap)
{
    char temp[100];
    int index = 0;
    int i = 0;
    for (; *s; s++) {
        if (isdigit(*s)) {
            temp[i++] = *s;
            // Need to avoid buffer overflow
            if (i == sizeof(temp)) {
                i = 0;
            }
        }
        if (isspace(*s)) {
            temp[i] = '\0';
            // strdup will allocate memory for the string, then copy it
            ar[index++] = strdup(temp);
            // if (NULL == ar[index-1]) TODO: Handle no memory error
            i = 0;
            if (index == MaxCap) {
                break;
            }
        }
    }
    if (index == 0) {
        printf("no numbers in string");
    }
    for (int i = 0; i < index; i++) {
        printf(" %s \n", ar[i]);
        // free the mem from strdup
        free(ar[i]);
    }
}
char * my_strdup(const char *src)
{
    if (src == NULL) return NULL;
    char *dest = malloc(strlen(src) + 1);
    if (dest == NULL) return NULL;
    strcpy(dest, src);
    return dest;
}
Trying to convert a String of digits to a array of numbers

Trying to convert a String of digits to a array of numbers


By : Ajay Jadhav
Date : March 29 2020, 07:55 AM
I wish did fix the issue. I'm not sure why have you tried so complicated stuff, but I think this might be useful:
I need to extract the first 6 digits and the last 4 digits of numbers in a string and delete leading zeros of the same w

I need to extract the first 6 digits and the last 4 digits of numbers in a string and delete leading zeros of the same w


By : Vickytech
Date : October 04 2020, 02:00 PM
wish helps you I need to delete leading zeros and extract the first 6 digits and the last 4 digits of numbers in a string with regular expression. , You could use this regex: /^0*(\d{6})\d*?(\d{4})$/
Explanation:
code :
function extractNumber(numberStr) {
  let match = numberStr.match(/^0*(\d{6})\d*?(\d{4})$/);
  return `${match[1]}${match[2]}`;
}

console.log(extractNumber('00005598001400248980'));
console.log(extractNumber('00000377820202008104'));
function extractNumber(numberStr, firstN = 6, lastN = 4) {
  let regex = new RegExp(`^0*(\\d{${firstN}})\\d*?(\\d{${lastN}})$`);
  let match = numberStr.match(regex);
  return `${match[1]}${match[2]}`;
}

console.log(extractNumber('00005598001400248980'));
console.log(extractNumber('00000377820202008104', 5));
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