I'm trying to take an integer as a parameter and then use recursion to double each digit in the integer.
For example doubleDigit(3487) would return 33448877.
I'm stuck because I can't figure out how I would read each number in the digit I guess.
To do this using recursion, use the modulus operator (%), dividing by 10 each time and accumulating your resulting string backwards, until you reach the base case (0), where there's nothing left to divide by. In the base case, you just return an empty string.
String doubleDigit(Integer digit) {
if (digit == 0) {
return "";
} else {
Integer thisDigit = digit % 10;
Integer remainingDigits = (digit - thisDigit) / 10;
return doubleDigit(remainingDigits) + thisDigit.toString() + thisDigit.toString();
}
}
If you're looking for a solution which returns an long instead of a String, you can use the following solution below (very similar to Chris', with the assumption of 0 as the base case):
long doubleDigit(long amt) {
if (amt == 0) return 0;
return doubleDigit(amt / 10) * 100 + (amt % 10) * 10 + amt % 10;
}
The function is of course limited by the maximum size of a long in Java.
I did the same question when doing Building Java Programs. Here is my solution which works for negative and positive numbers (and returns 0 for 0).
public static int doubleDigits(int n) {
if (n == 0) {
return 0;
} else {
int lastDigit = n % 10;
return 100 * doubleDigits(n / 10) + 10 * lastDigit + lastDigit;
}
There's no need to use recursion here.
I'm no longer a java guy, but an approximation of the algorithm I might use is this (works in C#, should translate directly to java):
int number = 3487;
int output = 0;
int shift = 1;
while (number > 0) {
int digit = number % 10; // get the least-significant digit
output += ((digit*10) + digit) * shift; // double it, shift it, add it to output
number /= 10; // move to the next digit
shift *= 100; // increase the amount we shift by two digits
}
This solution should work, but now that I've gone to the trouble of writing it, I realise that it is probably clearer to just convert the number to a string and manipulate that. Of course, that will be slower, but you almost certainly don't care about such a small speed difference :)
Edit:
Ok, so you have to use recursion. You already accepted a perfectly fine answer, but here's mine :)
private static long DoubleDigit(long input) {
if (input == 0) return 0; // don't recurse forever!
long digit = input % 10; // extract right-most digit
long doubled = (digit * 10) + digit; // "double" it
long remaining = input / 10; // extract the other digits
return doubled + 100*DoubleDigit(remaining); // recurse to get the result
}
Note I switched to long so it works with a few more digits.
You could get the String.valueOf(doubleDigit) representation of the given integer, then work with Commons StringUtils (easiest, in my opinion) to manipulate the String.
If you need to return another numeric value at that point (as opposed to the newly created/manipulated string) you can just do Integer.valueOf(yourString) or something like that.
Related
I'm trying to count the digits of an integer with java, it supposed to give 4 but instead I get 1 .
I don't know if I did something wrong, if someone could help.
Here's the code :
public static void main(String[] args) {
int n = 4781, i = 0;
while (n != 0) {
n %= 10;
n /= 10;
i++;
}
System.out.println(i);
}
The n %= 10 is unnecessary and in fact the source of the problem here. If you remove it, the code will work.
Usually when you do these kinds of "do something for each digit" you use % 10 to find the current digit a little bit like this:
int digit = n % 10;
System.out.println("The current digit is " + digit);
But in your case you don't actually care what the digits are, you only care how many there are, so there's no need to do % 10 at all.
And the way you did it you overwrote n (which is supposed to hold the current state of the number) with the current digit and then divided by 10. So no matter what the first digit is, this will never return a number bigger than 1.
This question already has answers here:
Java reverse an int value without using array
(33 answers)
Closed 3 years ago.
I'm a Java beginner so please pardon me if the question seems silly but I already searched the forums but it seems like no one has my problem.
I need to reverse the digits of an integer, and my class hasn't covered while or if loops yet, so I can't use those. All answers I can find on stackoverflow use those, so I can't use those.
the input I am given is below 10000 and above 0 and the code I have written has no problem reversing the integer if the input is 4 digits (e.g. 1000 - 9999) but once the input is between 1 - 999 it creates zeroes on the right hand side but according to the answer sheets its wrong.
For example: 1534 gets turned into 4351, but
403 becomes 3040 instead of the 304 it should be, and 4 becomes 4000 instead of 4.
I've tried different things in the code but it seems to just keep giving the same answer. Or maybe I'm just missing some key mathematics, I'm not sure.
Scanner scan = new Scanner(System.in);
System.out.println ("Enter an integer:");
int value = scan.nextInt();
int digit = (value % 10);
value = (value / 10);
int digit2 = (value % 10);
value = (value / 10);
int digit3 = (value % 10);
value = (value / 10);
int digit4 = (value % 10);
String reversednum = ("" + digit + digit2 + digit3 + digit4);
System.out.println ( reversednum);
and
Scanner scan = new Scanner(System.in);
System.out.println ("Enter an integer:");
int value = scan.nextInt();
int digit = (value % 10);
int reversednum = (digit);
value = (value /10);
digit = (value % 10);
reversednum = (reversednum * 10 + digit);
value = (value / 10);
digit = (value % 10);
reversednum = (reversednum * 10 + digit);
value = (value / 10);
digit = (value);
reversednum = (reversednum * 10 + digit);
System.out.println (reversednum);
What am I doing wrong?
You can convert from int to String -> reverse String -> convert again in int.
This is a code example.
public int getReverseInt(int value) {
String revertedStr = new StringBuilder(value).reverse().toString();
return Integer.parseInt(revertedStr);
}
Your code assumes that the number can be divided by 1000, which is clearly not the case for numbers below 1000. So add some if statements:
public int reverseNumber(int n) {
// step one: we find the factors using integer maths
int s = n;
int thousands = s / 1000; // this will be 0 if the number is <1000
s = s - thousands*1000;
int hundreds = s / 100; // this will be 0 if the number is <100
s = s - hundreds*100;
int tens = s / 10; // etc.
s = s - tens*10;
int ones = s;
// then: let's start reversing. single digit?
if (n<10) return n;
// two digits?
if (n<100) {
return ones*10 + tens;
}
// etc.
if (n<1000) {
return ones*100 + tens*10 + hundreds;
}
if (n<10000) {
return ones*1000 + tens*100 + hundreds*10 + thousands;
}
// if we get here, we have no idea what to do with this number.
return n;
}
Without spoon-feeding you code (leaving the value of writing your own homework code intact)...
Although you've said you can't use a loop, I don't think there's a sane approach that doesn't use one. Your basic problem is you have hard-coded a solution that works when the number happens to have 4 digits, rather than using code that adapts to a variable length. ie, are not using a loop.
All is not lost with your code however. You have figured out the essence of the solution. You just need to convert it to work processing one digit at a time. Consider using recursion, that divides the number by 10 each time and continues until the number is zero. Of course, you’ll have to capture the end digit before it’s lost by division.
Pseudo code may look like:
pass in the number and the current result
if the number is 0 return result
multiply result by 10 and add remainder of number divided by 10
return the result of calling self with number divided by 10 and result
then call this passing number and zero
Using modulus and division:
int nbr = 123; // reverse to 321 or 3*10*10 + 2*10 + 1
int rev = 0;
while(nbr > 0) {
rev *= 10; // shift left 1 digit
int temp = nbr % 10; // get LO digit
rev += temp; // add in next digit
nbr /= 10; // move to next digit
}
Or a recursive method:
public static int reverseInt(int number, int value) {
switch(number) { // is this conditional statement allowed???
case 0:
return value;
}
value *= 10;
int lod = number % 10;
value += lod;
number /= 10;
return reverseInt(number, value);
}
Suppose I have Long someLong = 1004L. What efficient method can I use to round this down to 1000L? Note that I do not actually know that someLong == 1004L so I can't simply do someLong -= 4L;. I need a generalizable method. I also want the ability to round down to each 5 instead of each 10, for example a function to round to 1005L (since if we're rounding by 5's then it'll round up instead of down).
More examples .. It could be that I have 1926L and I want to round to 5 meaning I need 1925L. Or I need to round to 10 meaning I need 1930L.
This is very simple.
If you want to round always down:
Your required formula is:
someLong-someLong%10
It is because someLong%10 is the remainder of someLong divided by 10. If you get this from the original number, you get the downrounded value, which you wanted.
The generalization is also simple: you can use 100, or even 13, if you want.
If you want to rounding in another direction (for example, rounding always up or always to the middle), then first to add something to this number, and then round always down.
If you want to round always up:
Then first you need to first add 9, then round always down.
someLong+9-(someLong+9)%10
If you want to round always to the middle:
...also you want to round to the nearest neightbor. Then you first add the half of the required interval, then round always down. For example, for 10 it is:
someLong+5-(someLong+5)%10
If you want to round a value towards the nearest multiple of step using the semantics of BigDecimal.ROUND_HALF_UP (if exactly halfway between two steps, round up), the necessary calculations are:
val += step/2;
val -= val%step;
Try this:
double a=1002l;
double b=a/10;
a=Math.round(b)*10;
System.out.println("Double round of value : "+a);
A generic function to round to the nearest multiple of k would be (works for positives only):
public static long round(long toRound, long k) {
long times = toRound / k;
long reminder = toRound % k;
if (reminder < k / 2) {
return k * times;
} else {
return k * (times + 1);
}
}
And a branchless variant (reminder < k / 2 => (2 * reminder / k) < 1:
public static long round(long toRound, long k) {
long times = toRound / k;
long reminder = toRound % k;
return k * (times + ((2 * reminder) / k));
}
The following example reachs what you need:
public static void main(String[] args) {
Long n = 1004L;
Long n2 = 1005L;
n = round(n);
n2 = round(n2);
System.out.println(n);
System.out.println(n2);
}
private static Long round(Long n) {
if (n%10 <=4) {
return n -=n%10;
} else {
return n += (10-n%10);
}
}
myFloor(long n, int m) {
return n - (n % m);
}
myRound(long n, int m) {
int i = (n % m) >= (m / 2) ? m : 0;
return n + i - (n % m);
}
so m could be 10 , 5 , ...
I have following code:
quesPart1 = ran.nextInt((numbersBetween - 2) + 1) + 2;
quesPart2 = ran.nextInt((numbersBetween - 2) + 1) + 2;
if(quesPart2 > quesPart1)
{
int placeHolder = quesPart1;
quesPart1 = quesPart2;
quesPart2 = placeHolder;
}
//if first part is even
if(quesPart1 % 2 == 0)
{
if(quesPart2 % 2 != 0)
{
--quesPart2;
}
}
else
{
if(quesPart2 % 2 == 0)
{
++quesPart2;
}
}
Above code make sure that if quesPart1 is greater than quesPart2 and both are even or both are odd numbers. Now i want to get only random numbers which are also divisible by one another. Like if i divide quesPart1 by quesPart2 i get integer not decimal number. Any ideas how i can do that without adding too much complexity to above code.
You can do something like:
int div = quesPart1 / quesPart2;
quesPart1 = div * quesPart2;
add this code at the bottom of your code.
Like if i divide quesPart1 by quesPart2 i get integer not decimal number.
Keep it simple: generate random numbers and take their product. Example:
quesPart2 = ran.nextInt(UPPER_BOUND);
int temp = ran.nextInt(UPPER_BOUND);
questPart1 = temp * quesPart2;
Specifying the range, as in the original question, is left an an exercise to the reader. (What, you didn't think I was going to do all the thinking for you, did you? ;-)
Look into the modulus operator, a % b. It returns the left over amount when a is divided by b. When b cleanly divides into a, such that there is no decimal part, a % b will be zero.
In order to generate a number that is divisible by another, given two random numbers, a and b, simply multiply a by b. This will give you c, a number that is a multiple of both a and b, and therefore dividable by both cleanly without remainder.
I have come up with this simple function and a do while loop that is easy to implement.
// This is a simple function to set the min and max integers you want
const getRandomIntInclusive = (min, max) => {
min = Math.ceil(min);
max = Math.floor(max);
return Math.floor(Math.random() * (max - min + 1)) + min;
}
//Define some variables variables
let firstNo = 0
let secondNo = 0
let isDivisible = 0;
//generate random ints until first number is divisible to second number
do {
//get random int between 1-9 for the first and second integer
firstNo = getRandomIntInclusive(1, 9)
secondNo = getRandomIntInclusive(1, 9)
isDivisible = firstNo % secondNo; //Check if it's fully divisible
}
while (isDivisible != 0) //Run until it is fully divisible
To generate Random numbers in java you can use ran.nextInt() or please refer to this link to see how to generate random numbers.
store those 2 random numbers (as num1 and num2).
To verify whether the solution after dividing num1 and num2 is integer or not, use this method:
sol = num1 / num2
if (sol == (int)sol)
{
... //true if the solution is an integer
}
How can i implement an algorithm to convert float or int to string?
I found one link
http://geeksforgeeks.org/forum/topic/amazon-interview-question-for-software-engineerdeveloper-0-2-years-about-algorithms-13
but i cant understand the algorithm given there
the numbers 0-9 are sequential in most character encoding so twiddling with the integral value of it will help here:
int val;
String str="";
while(val>0){
str = ('0'+(val%10)) + str;
val /= 10;
}
Here's a sample of how to do the integer to string, from it I hope you'll be able to figure out how to do the float to string.
public String intToString(int value) {
StringBuffer buffer = new StringBuffer();
if (value < 0) {
buffer.append("-");
}
// MAX_INT is just over 2 billion, so start by finding the number of billions.
int divisor = 1000000000;
while (divisor > 0) {
int digit = value / divisor; // integer division, so no remainder.
if (digit > 0) {
buffer.append('0'+digit);
value = value - digit * divisor; // subtract off the value to zero out that digit.
}
divisor = divisor / 10; // the next loop iteration should be in the 10's place to the right
}
}
This is of course, very unoptimized, but it gives you a feel for how the most basic formatting is accomplished.
Note that the technique of "" + x is actually rewritten to be something like
StringBuffer buffer = new StringBuffer();
buffer.append("");
buffer.append(String.valueOf(x));
buffer.toString();
So don't think that what is written is 100% exactly HOW it is done, look at is as what must happen in a larger view of things.
The general idea is to pick off the least significant digit by taking the number remainder ten. Then divide the number by 10 and repeat ... until you are left with zero.
Of course, it is a bit more complicated than that, especially in the float case.
if i have a single digit in int fomrat then i need to insert it into char , how to convert int to char?
Easy:
int digit = ... /* 0 to 9 */
char ch = (char)('0' + digit);
Well, you can read the code yourself.