Creating 2 numbers from a one #C - c#

I should create two new numbers from a one, the first group will contain digits which are divisible by 2 and the other group will contain the others.
int checkCount = 94321, num1 = 94321, count2 = 0, countRest = 0;
while (checkCount > 0)
{
if (checkCount % 2 == 0)
count2++;
else
countRest++;
checkCount /= 10;
}
int[] a = new int[count2];
int[] b = new int[countRest];
int k2 = 0, kRest = 0;
for (int j = 0; j < a.Length + b.Length; j++)
{
if (num1 % 2 == 0)
{
a[k2] = num1 % 10;
k2++;
}
else
{
b[kRest] = num1 % 10;
kRest++;
}
num1 /= 10;
}
I created two arrays with the numbers I should use, now how can I build two INT varabile when each one contains all of the numbers together from the array?
Example:
If I have this number - 12345 so
var = 24, other var = 135
If you have another solution without arrays I think it will be better.
Thank you.

Why not just:
int decimalMaskA = 1;
int decimalMaskB = 1;
while (checkCount > 0)
{
if (checkCount % 2 == 0)
{
count2 = count2 + (checkCount % 10)*decimalMaskA;
decimalMaskA *= 10;
}
else
{
countRest = countRest + (checkCount % 10)*decimalMaskB;
decimalMaskB *= 10;
}
checkCount /= 10;
}
count2 and countRest will contain those numbers (135 and 24) instead of counts.

This splits number 12345 to numbers 135 and 24.
int checkCount = 12345;
int even = 0;
int odd = 0;
int reverseEven = 0;
int reverseOdd = 0;
while (checkCount > 0) {
int current = checkCount % 10;
if (current % 2 == 0) {
reverseEven = 10 * reverseEven + current;
} else {
reverseOdd = 10 * reverseOdd + current;
}
checkCount /= 10;
}
while (reverseEven > 0) {
even = 10 * even + reverseEven % 10;
reverseEven /= 10;
}
while (reverseOdd > 0) {
odd = 10 * odd + reverseOdd % 10;
reverseOdd /= 10;
}
Console.WriteLine("even: {0}", even);
Console.WriteLine("odd: {0}", odd);

If I understand what you're looking for this will do the trick:
//Setup a sample array
int[] a = new int[2];
a[0] = 2;
a[1] = 4;
//Convert each item to a string, convert that to a string array, join the strings and turn into an int
int output = int.Parse(String.Join("", a.Select(s => s.ToString()).ToArray()));

This works for me:
int number = 12345;
string result1 = "";
string result2 = "";
string numberString = number.ToString();
for (int i = 0; i < numberString.Length; i++ )
{
if (numberString[i] % 2 == 0)
{
result1 = result1 + numberString[i];
}
else
{
result2 = result2 + numberString[i];
}
}
int evenNumbers = int.Parse(result1);
int oddNumbers = int.Parse(result2);

How can I build two INT varabile
when each one contains all of the
numbers together from the array?
I can't say for sure, but I think you're asking how to assemble a number given each of its digits in decreasing order of significance.
To 'append' a digit to a number, you can multiply the number by 10 and then add the digit to that. To create the 'assembled' number, you can perform this operation for each digit in the array,
int[] digits = ...
int num = digits.Aggregate(0, (numSoFar, digit) => 10 * numSoFar + digit);
As a loop, this would look like:
int num = 0;
foreach(int digit in digits)
{
num = 10 * num + digit;
}

Try this with LINQ,
int num = 92345;
string strNum = Convert.ToString(num);
var divisibleby2 = from c in strNum
where int.Parse(c.ToString()) % 2 == 0
select c.ToString();
var notDivisibleby2 = from c in strNum
where int.Parse(c.ToString()) % 2 != 0
select c.ToString();
int int_divisibleby2num = int.Parse(String.Join("", divisibleby2.ToArray()));
int int_Notdivisibleby2num = int.Parse(String.Join("", notDivisibleby2.ToArray()));

Every programmer should write wacky code at least once a day:
int checkCount = 12345, numEven, numOdd;
Boolean result;
result = int.TryParse(checkCount.ToString().Replace("0", "").Replace("2", "").Replace("4", "").Replace("6", "").Replace("8", ""), out numOdd);
result = int.TryParse(checkCount.ToString().Replace("1", "").Replace("3", "").Replace("5", "").Replace("7", "").Replace("9", ""), out numEven);

Another solution could be...
var num1 = 94321;
var oddFinal = 0;
var evenFinal = 0;
var odd = new List<int>();
var even = new List<int>();
while( num1>0 )
{
if( num1 % 2 == 0 )
odd.Add( num1 % 10 );
else
even.Add( num1 % 10 );
num1 = num1 / 10;
}
for (int i = 0; i < odd.Count; i++)
{
oddFinal += odd[i] * (int) Math.Pow(10,i);
}
for (int i = 0; i < even.Count; i++)
{
evenFinal += even[i] * (int) Math.Pow(10,i);
}

Related

inaccurate results with function to add an array of digits together

so i have this function:
static int[] AddArrays(int[] a, int[] b)
{
int length1 = a.Length;
int length2 = b.Length;
int carry = 0;
int max_length = Math.Max(length1, length2) + 1;
int[] minimum_arr = new int[max_length - length1].Concat(a).ToArray();
int[] maximum_arr = new int[max_length - length2].Concat(b).ToArray();
int[] new_arr = new int[max_length];
for (int i = max_length - 1; i >= 0; i--)
{
int first_digit = maximum_arr[i];
int second_digit = i - (max_length - minimum_arr.Length) >= 0 ? minimum_arr[i - (max_length - minimum_arr.Length)] : 0;
if (second_digit + first_digit + carry > 9)
{
new_arr[i] = (second_digit + first_digit + carry) % 10;
carry = 1;
}
else
{
new_arr[i] = second_digit + first_digit + carry;
carry = 0;
}
}
if (carry == 1)
{
int[] result = new int[max_length + 1];
result[0] = 1;
Array.Copy(new_arr, 0, result, 1, max_length);
return result;
}
else
{
return new_arr;
}
}
it basically takes 2 lists of digits and adds them together. the point of this is that each array of digits represent a number that is bigger then the integer limits. now this function is close to working the results get innacurate at certein places and i honestly have no idea why. for example if the function is given these inputs:
"1481298410984109284109481491284901249018490849081048914820948019" and
"3475893498573573849739857349873498739487598" (both of these are being turned into a array of integers before being sent to the function)
the expected output is:
1,481,298,410,984,109,284,112,957,384,783,474,822,868,230,706,430,922,413,560,435,617
and what i get is:
1,481,298,410,984,109,284,457,070,841,142,258,634,158,894,233,092,241,356,043,561,7
i would very much appreciate some help with this ive been trying to figure it out for hours and i cant seem to get it to work perfectly.
I suggest Reverse arrays a and b and use good old school algorithm:
static int[] AddArrays(int[] a, int[] b) {
Array.Reverse(a);
Array.Reverse(b);
int[] result = new int[Math.Max(a.Length, b.Length) + 1];
int carry = 0;
int value = 0;
for (int i = 0; i < Math.Max(a.Length, b.Length); ++i) {
value = (i < a.Length ? a[i] : 0) + (i < b.Length ? b[i] : 0) + carry;
result[i] = value % 10;
carry = value / 10;
}
if (carry > 0)
result[result.Length - 1] = carry;
else
Array.Resize(ref result, result.Length - 1);
// Let's restore a and b
Array.Reverse(a);
Array.Reverse(b);
Array.Reverse(result);
return result;
}
Demo:
string a = "1481298410984109284109481491284901249018490849081048914820948019";
string b = "3475893498573573849739857349873498739487598";
string c = string.Concat(AddArrays(
a.Select(d => d - '0').ToArray(),
b.Select(d => d - '0').ToArray()));
Console.Write(c);
Output:
1481298410984109284112957384783474822868230706430922413560435617

How to get integers to an array (from integer i to integer j). c#

I want to store integers(in an array or anything) that in range of int "i" and int "j".
eg:-Think, "int i = 1" and "int j = 10".I want to store integers from 1 and 10.
So that (1,2,3,4,5,6,7,8,9,10)
Because I want to answer to HackerRank "Beautiful Days at the Movies".
link below.
https://www.hackerrank.com/challenges/beautiful-days-at-the-movies/problem?isFullScreen=false
here is my code and it a garbage.
static int beautifulDays(int i, int j, int k) {
var total = 0;
for(var a = i; a <= j; a++ )
{
if (a != 0)
{
int ri = Reverse(i);
int rj = Reverse(j);
var ra = Reverse(a);
if((ra/k) % 1 == 0)
{
total++;
}
if((rj/k) % 1 == 0)
{
total++;
}
if((ri/k) % 1 == 0)
{
total++;
}
}
return total;
}
return total;
}
public static int Reverse(int inval)
{
int result = 0;
do
{
result = (result * 10) + (inval % 10);
inval = inval / 10;
}
while(inval > 0);
return result;
}
simply, can you give me the answer of HackerRank "Beautiful Days at the Movies".
link below.
https://www.hackerrank.com/challenges/beautiful-days-at-the-movies/problem?isFullScreen=false
Using Java you can easily stream a range of numbers with IntStream, then map the reverse function for each value, then filter those that fulfils the condition and count. With streams you don't need to store, you can get straight to the answer.
IntUnaryOperator reverse = (opperand) -> {
int reversed = 0;
int num = opperand;
while (num != 0) {
int digit = num % 10;
reversed = reversed * 10 + digit;
num /= 10;
}
return Math.abs(opperand - reversed);
};
return (int) IntStream.rangeClosed(i, j).map(reverse)
.filter(v -> v % k == 0).count();

How to loop random number to make result equal user input summary [duplicate]

This question already has answers here:
How to generate random 5 digit number depend on user summary
(2 answers)
Closed 4 years ago.
i have some problem like this.
user(A) enter 500000 to my application and then i want to generate 50 random 5 digit number and when summary 50 random number need to equal 500000, How to do like i tried already but it's not working this is my code
int balane = 500000;
int nums = 50;
int max = balane / nums;
Random rand = new Random();
int newNum = 0;
int[] ar = new int[nums];
for (int i = 0; i < nums - 1; i++)
{
newNum = rand.Next(max);
ar[i] = newNum;
balane -= newNum;
max = balane / (nums - i - 1);
ar[nums - 1] = balane;
}
int check = 0;
foreach (int x in ar)
{
check += x;
}
the result that i tell you not working because in my array list i have value more than 5 digit but the result equal 500000.
How to solve this issue ? Thank you very much.
This works for me
public static void test(int balance = 500000, int nums = 50, int max_digits = 5)
{
int rest = balance;
var max_value = (int)Math.Pow(10, max_digits) - 1;
var rand = new Random();
int[] ar = new int[nums];
for (int i = 0; i < nums - 1; i++)
{
var max = rest / (nums - i);
if (max > max_value) max = max_value;
var newNum = rand.Next(max);
ar[i] = newNum;
rest -= newNum;
}
while (rest > max_value)
{
var all_values_are_max = true;
for (int i = 0; i < nums - 1; i++)
{
if (ar[i] < max_value)
{
var d = (int)((max_value - ar[i]) / 10.0); // 10% increase
ar[i] += d;
rest -= d;
all_values_are_max = false;
}
}
if (all_values_are_max)
throw new Exception("This is not possible at all!");
}
while (rest < 0)
{
for (int i = 0; i < nums - 1; i++)
{
if (ar[i] > 0)
{
var d = (int)(ar[i] / 20.0); // %5 decrease
ar[i] -= d;
rest += d;
}
}
}
ar[nums - 1] = rest;
int check_sum = 0;
foreach (int x in ar)
{
check_sum += x;
if (x > max_value || x < 0)
MessageBox.Show("wrong value");
}
if (check_sum != balance)
MessageBox.Show("Error: sum is " + check_sum);
MessageBox.Show("ok");
}
For example, test(500000,50) works fine all times, but test(500000, 5) throw an exception, because is not possible
Perhaps try this:
var rnd = new Random();
var numbers = Enumerable.Range(0, 50).Select(x => rnd.Next(500_000)).OrderBy(x => x).ToArray();
numbers = numbers.Skip(1).Zip(numbers, (x1, x0) => x1 - x0).ToArray();
numbers = numbers.Append(500_000 - numbers.Sum()).ToArray();
Console.WriteLine(numbers.Count());
Console.WriteLine(numbers.Sum());
This outputs:
50
500000
This works by generating 50 random numbers between 0 and 499,999 inclusively. It then sorts them ascendingly and then gets the difference between each successive pair. This by definition produces a set of 49 values that almost adds up to 500,000. It's then just a matter of adding the one missing number by doing 500_000 - numbers.Sum().

Reorder digits in integer using C#

I want to ask how I can reorder the digits in an Int32 so they result in the biggest possible number.
Here is an example which visualizes what I am trying to do:
2927466 -> 9766422
12492771 -> 97742211
I want to perform the ordering of the digits without using the System.Linq namespace and without converting the integer into a string value.
This is what I got so far:
public static int ReorderInt32Digits(int v)
{
int n = Math.Abs(v);
int l = ((int)Math.Log10(n > 0 ? n : 1)) + 1;
int[] d = new int[l];
for (int i = 0; i < l; i++)
{
d[(l - i) - 1] = n % 10;
n /= 10;
}
if (v < 0)
d[0] *= -1;
Array.Sort(d);
Array.Reverse(d);
int h = 0;
for (int i = 0; i < d.Length; i++)
{
int index = d.Length - i - 1;
h += ((int)Math.Pow(10, index)) * d[i];
}
return h;
}
This algorithm works flawlessly but I think it is not very efficient.
I would like to know if there is a way to do the same thing more efficiently and how I could improve my algorithm.
You can use this code:
var digit = 2927466;
String.Join("", digit.ToString().ToCharArray().OrderBy(x => x));
Or
var res = String.Join("", digit.ToString().ToCharArray().OrderByDescending(x => x) );
Not that my answer may or may not be more "efficient", but when I read your code you calculated how many digits there are in your number so you can determine how large to make your array, and then you calculated how to turn your array back into a sorted integer.
It would seem to me that you would want to write your own code that did the sorting part without using built in functionality, which is what my sample does. Plus, I've added the ability to sort in ascending or descending order, which is easy to add in your code too.
UPDATED
The original algorithm sorted the digits, now it sorts the digits so that the end result is the largest or smallest depending on the second parameter passed in. However, when dealing with a negative number the second parameter is treated as opposite.
using System;
public class Program
{
public static void Main()
{
int number1 = 2927466;
int number2 = 12492771;
int number3 = -39284925;
Console.WriteLine(OrderDigits(number1, false));
Console.WriteLine(OrderDigits(number2, true));
Console.WriteLine(OrderDigits(number3, false));
}
private static int OrderDigits(int number, bool asc)
{
// Extract each digit into an array
int[] digits = new int[(int)Math.Floor(Math.Log10(Math.Abs(number)) + 1)];
for (int i = 0; i < digits.Length; i++)
{
digits[i] = number % 10;
number /= 10;
}
// Order the digits
for (int i = 0; i < digits.Length; i++)
{
for (int j = i + 1; j < digits.Length; j++)
{
if ((!asc && digits[j] > digits[i]) ||
(asc && digits[j] < digits[i]))
{
int temp = digits[i];
digits[i] = digits[j];
digits[j] = temp;
}
}
}
// Turn the array of digits back into an integer
int result = 0;
for (int i = digits.Length - 1; i >= 0; i--)
{
result += digits[i] * (int)Math.Pow(10, digits.Length - 1 - i);
}
return result;
}
}
Results:
9766422
11224779
-22345899
See working example here... https://dotnetfiddle.net/RWA4XV
public static int ReorderInt32Digits(int v)
{
var nums = Math.Abs(v).ToString().ToCharArray();
Array.Sort(nums);
bool neg = (v < 0);
if(!neg)
{
Array.Reverse(nums);
}
return int.Parse(new string(nums)) * (neg ? -1 : 1);
}
This code fragment below extracts the digits from variable v. You can modify it to store the digits in an array and sort/reverse.
int v = 2345;
while (v > 0) {
int digit = v % 10;
v = v / 10;
Console.WriteLine(digit);
}
You can use similar logic to reconstruct the number from (sorted) digits: Multiply by 10 and add next digit.
I'm posting this second answer because I think I got the most efficient algorithm of all (thanks for the help Atul) :)
void Main()
{
Console.WriteLine (ReorderInt32Digits2(2927466));
Console.WriteLine (ReorderInt32Digits2(12492771));
Console.WriteLine (ReorderInt32Digits2(-1024));
}
public static int ReorderInt32Digits2(int v)
{
bool neg = (v < 0);
int mult = neg ? -1 : 1;
int result = 0;
var counts = GetDigitCounts(v);
for (int i = 0; i < 10; i++)
{
int idx = neg ? 9 - i : i;
for (int j = 0; j < counts[idx]; j++)
{
result += idx * mult;
mult *= 10;
}
}
return result;
}
// From Atul Sikaria's answer
public static int[] GetDigitCounts(int n)
{
int v = Math.Abs(n);
var result = new int[10];
while (v > 0) {
int digit = v % 10;
v = v / 10;
result[digit]++;
}
return result;
}

Integer to Integer Array C# [duplicate]

This question already has answers here:
Is there an easy way to turn an int into an array of ints of each digit?
(11 answers)
Closed 1 year ago.
I had to split an int "123456" each value of it to an Int[] and i have already a Solution but i dont know is there any better way :
My solution was :
public static int[] intToArray(int num){
String holder = num.ToString();
int[] numbers = new int[Holder.ToString().Length];
for(int i=0;i<numbers.length;i++){
numbers[i] = Convert.toInt32(holder.CharAt(i));
}
return numbers;
}
A simple solution using LINQ
int[] result = yourInt.ToString().Select(o=> Convert.ToInt32(o) - 48 ).ToArray()
I believe this will be better than converting back and forth. As opposed to JBSnorro´s answer I reverse after converting to an array and therefore avoid IEnumerable´s which I think will contribute to a little bit faster code. This method work for non negative numbers, so 0 will return new int[1] { 0 }.
If it should work for negative numbers, you could do a n = Math.Abs(n) but I don't think that makes sense.
Furthermore, if it should be more performant, I could create the final array to begin with by making a binary-search like combination of if-statements to determine the number of digits.
public static int[] digitArr(int n)
{
if (n == 0) return new int[1] { 0 };
var digits = new List<int>();
for (; n != 0; n /= 10)
digits.Add(n % 10);
var arr = digits.ToArray();
Array.Reverse(arr);
return arr;
}
Update 2018:
public static int numDigits(int n) {
if (n < 0) {
n = (n == Int32.MinValue) ? Int32.MaxValue : -n;
}
if (n < 10) return 1;
if (n < 100) return 2;
if (n < 1000) return 3;
if (n < 10000) return 4;
if (n < 100000) return 5;
if (n < 1000000) return 6;
if (n < 10000000) return 7;
if (n < 100000000) return 8;
if (n < 1000000000) return 9;
return 10;
}
public static int[] digitArr2(int n)
{
var result = new int[numDigits(n)];
for (int i = result.Length - 1; i >= 0; i--) {
result[i] = n % 10;
n /= 10;
}
return result;
}
int[] outarry = Array.ConvertAll(num.ToString().ToArray(), x=>(int)x);
but if you want to convert it to 1,2,3,4,5:
int[] outarry = Array.ConvertAll(num.ToString().ToArray(), x=>(int)x - 48);
I'd do it like this:
var result = new List<int>();
while (num != 0) {
result.Insert(0, num % 10);
num = num / 10;
}
return result.ToArray();
Slightly less performant but possibly more elegant is:
return num.ToString().Select(c => Convert.ToInt32(c.ToString())).ToArray();
Note that these both return 1,2,3,4,5,6 rather than 49,50,51,52,53,54 (i.e. the byte codes for the characters '1','2','3','4','5','6') as your code does. I assume this is the actual intent?
Using conversion from int to string and back probably isn't that fast. I would use the following
public static int[] ToDigitArray(int i)
{
List<int> result = new List<int>();
while (i != 0)
{
result.Add(i % 10);
i /= 10;
}
return result.Reverse().ToArray();
}
I do have to note that this only works for strictly positive integers.
EDIT:
I came up with an alternative. If performance really is an issue, this will probably be faster, although you can only be sure by checking it yourself for your specific usage and application.
public static int[] ToDigitArray(int n)
{
int[] result = new int[GetDigitArrayLength(n)];
for (int i = 0; i < result.Length; i++)
{
result[result.Length - i - 1] = n % 10;
n /= 10;
}
return result;
}
private static int GetDigitArrayLength(int n)
{
if (n == 0)
return 1;
return 1 + (int)Math.Log10(n);
}
This works when n is nonnegative.
Thanks to ASCII character table. The simple answer using LINQ above yields answer + 48.
Either
int[] result = youtInt.ToString().Select(o => Convert.ToInt32(o) - 48).ToArray();
or
int[] result = youtInt.ToString().Select(o => int.Parse(o.ToString())).ToArray();
can be used
You can do that without converting it to a string and back:
public static int[] intToArray(int num) {
List<int> numbers = new List<int>();
do {
numbers.Insert(0, num % 10);
num /= 10;
} while (num > 0);
return numbers.ToArray();
}
It only works for positive values, of course, but your original code also have that limitation.
I would convert it in the below manner
if (num == 0) return new int[1] { 0 };
var digits = new List<int>();
while (num > 0)
{
digits.Add(num % 10);
num /= 10;
}
var arr = digits.ToArray().Reverse().ToArray();
string DecimalToBase(int iDec, int numbase)
{
string strBin = "";
int[] result = new int[32];
int MaxBit = 32;
for(; iDec > 0; iDec/=numbase)
{
int rem = iDec % numbase;
result[--MaxBit] = rem;
}
for (int i=0;i<result.Length;i++)
if ((int)result.GetValue(i) >= base10)
strBin += cHexa[(int)result.GetValue(i)%base10];
else
strBin += result.GetValue(i);
strBin = strBin.TrimStart(new char[] {'0'});
return strBin;
}
int BaseToDecimal(string sBase, int numbase)
{
int dec = 0;
int b;
int iProduct=1;
string sHexa = "";
if (numbase > base10)
for (int i=0;i<cHexa.Length;i++)
sHexa += cHexa.GetValue(i).ToString();
for(int i=sBase.Length-1; i>=0; i--,iProduct *= numbase)
{
string sValue = sBase[i].ToString();
if (sValue.IndexOfAny(cHexa) >=0)
b=iHexaNumeric[sHexa.IndexOf(sBase[i])];
else
b= (int) sBase[i] - asciiDiff;
dec += (b * iProduct);
}
return dec;
}
i had similar requirement .. i took from many good ideas, and added a couple missing pieces .. where many folks weren’t handling zero or negative values. this is what i came up with:
public static int[] DigitsFromInteger(int n)
{
int _n = Math.Abs(n);
int length = ((int)Math.Log10(_n > 0 ? _n : 1)) + 1;
int[] digits = new int[length];
for (int i = 0; i < length; i++)
{
digits[(length - i) - 1] = _n % 10 * ((i == (length - 1) && n < 0) ? -1 : 1);
_n /= 10;
}
return digits;
}
i think this is pretty clean .. although, it is true we're doing a conditional check and several extraneous calculations with each iteration .. while i think they’re nominal in this case, you could optimize a step further this way:
public static int[] DigitsFromInteger(int n)
{
int _n = Math.Abs(n);
int length = ((int)Math.Log10(_n > 0 ? _n : 1)) + 1;
int[] digits = new int[length];
for (int i = 0; i < length; i++)
{
//digits[(length - i) - 1] = _n % 10 * ((i == (length - 1) && n < 0) ? -1 : 1);
digits[(length - i) - 1] = _n % 10;
_n /= 10;
}
if (n < 0)
digits[0] *= -1;
return digits;
}
A slightly more concise way to do MarkXA's one-line version:
int[] result = n.ToString().Select(c => (int)Char.GetNumericValue(c)).ToArray();
GetNumericValue returns the visible number in your char as a double, so if your string is "12345", it will return the doubles 1,2,3,4,5, which can each be cast to an int. Note that using Convert.ToInt32 on a char in C# returns the ASCII code, so you would get 49,50,51,52,53. This can understandably lead to a mistake.
Here is a Good Solution for Convert Your Integer into Array i.e:
int a= 5478 into int[]
There is no issue if You Have a String and You want to convert a String into integer Array for example
string str=4561; //Convert into
array[0]=4;
array[1]=5;
array[2]=6;
array[3]=7;
Note: The Number of zero (0) in devider are Equal to the Length of input and Set Your Array Length According to Your input length
Now Check the Coding:
string str=4587;
int value = Convert.ToInt32(str);
int[] arr = new int[4];
int devider = 10000;
for (int i = 0; i < str.Length; i++)
{
int m = 0;
devider /= 10;
arr[i] = value / devider;
m = value / devider;
value -= (m * devider);
}
private static int[] ConvertIntToArray(int variable)
{
string converter = "" + variable;
int[] convertedArray = new int[converter.Length];
for (int i=0; i < convertedArray.Length;i++) //it can be also converter.Length
{
convertedArray[i] = int.Parse(converter.Substring(i, 1));
}
return convertedArray;
}
We get int via using method. Then, convert it to string immediately (123456->"123456"). We have a string called converter and carry to int value. Our string have a string.Length, especially same length of int so, we create an array called convertedArray that we have the length, that is converter(string) length. Then, we get in the loop where we are convert the string to int one by one via using string.Substring(i,1), and assign the value convertedArray[i]. Then, return the convertedArray.At the main or any method you can easily call the method.
public static int[] intToArray(int num)
{
num = Math.Abs(num);
int length = num.ToString().Length;
int[] arr = new int[length];
do
{
arr[--length] = num % 10;
num /= 10;
} while (num != 0);
return arr;
}
Dividing by system base (decimal in this case) removes the right most digit, and we get that digit by remainder operator. We keep repeating until we end up with a zero. Each time a digit is removed it will be stored in an array starting from the end of the array and backward to avoid the need of revering the array at the end. The Math.Abs() function is to handle the negative input, also the array is instantiated with the same size as input length.
Integer or Long to Integer Array C#
Convert it to char array and subtract 48.
public static int[] IntToArray(int value, int length)
{
char[] charArray = new char[length];
charArray = value.ToString().ToCharArray();
int[] intArray = new int[length];
for (int i = 0; i < intArray.Length; i++)
{
intArray[i] = charArray[i] - 48;
}
return intArray;
}
public static int[] LongToIntArray(long value, int length)
{
char[] charArray = new char[length];
charArray = value.ToString().ToCharArray();
int[] intArray = new int[length];
for (int i = 0; i < intArray.Length; i++)
{
intArray[i] = charArray[i] - 48;
}
return intArray;
}

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