I want to convert the decimal number into a binary number using recursion in java. I tried a lot but unable to do it. Here is my code:
public class DecimalToBinary {
public static void main(String[] args) {
System.out.println(conversion(2));
}
public static int conversion(int n) {
return reconversion(n);
}
public static int reconversion(int n) {
if(n <= 0)
return 0;
else {
return (int) (n/2 + conversion(n/2));
}
}
}
Integer values are already in binary. The fact that they appear as digits 0 thru 9 when you print them is because they are converted to a string of decimal digits. So you need to return a String of binary digits like so.
public static String conversion(int n) {
String b = "";
if (n > 1) {
// continue shifting until n == 1
b = conversion(n >> 1);
}
// now concatenate the return values based on the logical AND
b += (n & 1);
return b;
}
Related
public class linklist {
public static void main(String[] args) {
int a = 00486;
int x=zero(a);
System.out.println(x);
}
public static int zero(int n)
{
if(n<=10)
{
return 0;
}
if(n%10==0) {
return 1 + zero(n / 10);
}
else
return zero(n/10);
}
}
in line 3 i set 00486 as value but its showing integer too large error.As per my knowledge In Java, the integer value permissible is much bigger.
When you add 0 in front of the number, it is treated in the octal format. In octal format, the allowed digits vary from 0 to 7 only .
Therefore, you need to change the number from 00486 to 00476. But be beware, this number will be converted to base 10 format. To verify that, I have written a print statement that shows that the number will be stored indeed in base 10 format.
public class linklist {
public static void main(String[] args) {
int a = 0476;
System.out.println(8*8*4 + 8*7 + 6);
System.out.println(a);
int x=zero(a);
System.out.println(x);
}
public static int zero(double n)
{
if(n<=10)
{
return 0;
}
if(n%10==0) {
return 1 + zero(n / 10);
}
else
return zero(n/10);
}
}
I want to reverse an int but it doesn't work. For example, 123 should return 321, but the printed number is 356.
public class x {
public static void main(String[] args) {
System.out.println(reverse2(123, 0));
}
static int reverse2(int a, int i) {
if(a == 0) {
return 0;
} else {
i = i*10 + a%10;
System.out.println(i);
return i += reverse2(a/10, i);
}
}
}
Your code should look like this:
public class x {
public static void main(String[] args) {
System.out.println(reverse2(123, 0));
}
static int reverse2(int a, int i) {
if(a == 0) {
return i;
} else {
i = i*10 + a%10;
System.out.println(i);
return reverse2(a/10, i);
}
}
}
You should return i when a is 0.
You shouldn't add i when you call the reverse2 function because you're adding i twice.
You are greatly complicating your recursive function for printing an integer in reverse. For one, there is no good reason for reverse2 to have two integer arguments, as you can achieve your desired results with a single argument. The trick is to access the rightmost digit with the % 10 operation then shift that digit off the number with the / 10 operation. Consider these revisions:
public class x {
public static void main(String[] args) {
System.out.println(reverse2(123));
}
static String reverse2(int number) {
if(number == 0) {
return "";
} else {
return number % 10 + reverse2(number / 10);
}
}
}
You can do it like this. You only need to pass the value you are reversing. The math computation computes 10 to the power of the number of digits in the argument.
public static int reverse(int v) {
int reversed = 0;
if (v > 0) {
int d = (int)Math.pow(10,(int)(Math.log10(v)));
reversed = reverse(v%d) * 10 + v/d;
}
return reversed;
}
Of course, if you can pass a second argument as a scratch pad, then it can be done like so. As you tear down the original value you build up the returned value.
public static int reverse(int v, int reversed) {
if (v > 0) {
return reverse(v / 10, reversed * 10 + v % 10);
}
return reversed;
}
Problem Statement: Given a 32-bit signed integer, reverse digits of an integer.
Example 1:
Input: 123
Output: 321
Example 2:
Input: -123
Output: -321
My Solution:
class Solution7{
public int reverse(int x) {
if(x>Integer.MAX_VALUE || x<Integer.MIN_VALUE) {
return 0;
}
StringBuilder S_rev = new StringBuilder();
String S_r_v=S_rev.append(Math.abs(x)).reverse().toString();//.toString() String builder to String
double reverse_no=Double.parseDouble(S_r_v);
if (x < 0) {
return -(int)reverse_no;
}
return (int)reverse_no;
}
}
My Solution is ok for most of the test case. But it cannot pass one test case and I got a error
Error: Line 10: java.lang.NumberFormatException: For input string: "8463847412-"
If someone know what type of error it is please discuss.
Thank you in advance.
It seems like you are trying to pass in Integer.MIN_VALUE
When you pass in the minimum integer value, Math.abs seems to return a negative number as stated here
https://docs.oracle.com/javase/8/docs/api/java/lang/Math.html#abs-int-
Note that if the argument is equal to the value of Integer.MIN_VALUE, the most negative representable int value, the result is that same value, which is negative.
You can either check for x<=Integer.MIN_VALUE and return 0 if x is Integer.MIN_VALUE or handle the special case for Integer.MIN_VALUE
if(x== Integer.MIN_VALUE)
return -8463847412;
By converting number to String and reversing the sign symbol ended up on the end of the value. This makes the number invalid.
You don't have to convert to String or double. You can use module operator % to extract digits:
public int reverse(int x) {
long result = 0;
while (x != 0) {
result *= 10;
result += x % 10;
x /= 10;
}
if (result > Integer.MAX_VALUE || result < Integer.MIN_VALUE) {
throw new IllegalArgumentException(); // overflow
}
return result;
}
If you necessarily want to implement it using StringBuilder, here it is:
public static void main(String[] args) {
ReverseNum reverseNum = new ReverseNum();
System.out.println(reverseNum.reverse(-123));
System.out.println(reverseNum.reverse(123));
System.out.println(reverseNum.reverse(0));
}
public int reverse(int x) {
int res = 1;
String xStr = String.valueOf(x);
StringBuilder builder = null;
if (xStr.startsWith("-")) {
builder = new StringBuilder(xStr.substring(1));
res = -1;
} else {
builder = new StringBuilder(xStr);
}
return res * Integer.valueOf(builder.reverse().toString());
}
Output:
-321
321
0
P.S. If you want to avoid integer overflow, then you can simply use long instead of int, like this:
public long reverse(int x) {
long res = 1;
String xStr = String.valueOf(x);
StringBuilder builder = null;
if (xStr.startsWith("-")) {
builder = new StringBuilder(xStr.substring(1));
res = -1;
} else {
builder = new StringBuilder(xStr);
}
return res * Long.valueOf(builder.reverse().toString());
}
public class ReverseString {
public static void main(String args[]) {
ReverseString rs = new ReverseString();
System.out.println(rs.reverse(-84638));
System.out.println(rs.reverse(5464867));
}
public long reverse(int number) {
boolean isNegative = number < 0;
StringBuilder reverseBuilder = new StringBuilder();
String reversedString = reverseBuilder.append(Math.abs(number)).reverse().toString();
long reversedStringValue = Long.parseLong(reversedString);
if(isNegative) {
return reversedStringValue * -1;
} else {
return reversedStringValue;
}
}
}
This code provides the output you have mentioned in the requirement. And It also supports for integer overflow. Your requirement is to convert int values. It is okay to get the converted value in the higher format since converted value may not be in the range of int. I have changed the reverse method return type to long.
I have identified a few issues in your code.
public int reverse(int x) {
if(x>Integer.MAX_VALUE || x<Integer.MIN_VALUE) {
return 0;
}
Above code segment, not point of checking whether the value is inside the int range because it is already received in the param as a string. It should throw an error before executing your code lines since it is not able to fit the larger value to int variable.
Finally, the int number you have used is not in the int range. (-8463847412)
What about this?
public class ReverseNumber {
public static void main(String[] args) {
System.out.println(reverse(123456));
System.out.println(reverse(0));
System.out.println(reverse(-987654));
}
private static int reverse(int i) {
final int signum;
if(i < 0) {
signum = -1;
} else {
signum = +1;
}
int reversedNumber = 0;
int current = Math.abs(i);
while(0 < current) {
final int cipher = current % 10;
reversedNumber = Math.addExact(Math.multiplyExact(reversedNumber, 10), cipher);
current = current / 10;
}
return signum * reversedNumber;
}
}
Output:
654321
0
-456789
This solution avoids strings and can handle negative numbers.
It throws an Arithmetic exception if an integer overflow happens.
The input for this method is "9876548"
It returns
8
4
5
6
7
8
9
9876548
I don't want the "9876548" at the end.
(Stack over flow format wont all
Implement a recursive method printDigits that takes an integer num as a parameter and prints its digits in reverse order, one digit per line.
public class PrintDigits{
public static void main (String [] args)
{System.out.print(reversDigits(9876548));}
public static int reversDigits(int number) {
int result;
if (number < 10) {
System.out.println(number);
result = number;
}
else{
System.out.println(number % 10);
reversDigits(number/10);
result = number;
}
return result;
}
}
Thank you for your help!
Change this
System.out.print(reversDigits(9876548));
to
reversDigits(9876548);
Use this way
public static void main (String [] args)
{
reversDigits(9876548);
}
public static int reversDigits(int number) {
int result;
if (number < 10) {
System.out.println(number);
result = number;
}
else{
System.out.println(number % 10);
reversDigits(number/10);
result = number;
}
return result;
}
I am practicing recursion and I can't see why this method does not seem to work.
Any ideas?
public void fact()
{
fact(5);
}
public int fact(int n)
{
if(n == 1){
return 1;
}
return n * (fact(n-1));
}
}
Thanks
Your code seems to work but you are not doing anything with the returned value, put method call fact or fact(5) inside of a System.out.println and see what you get.
The recursion part is fine; you're just not using its return value, which gets discarded. Here's a complete Java application of your factorial code, slightly jazzed-up for educational purposes:
public class Factorial {
public static String fact(int n) {
if(n == 1){
return "1";
}
return n + " * " + (fact(n-1)); // what happens if you switch the order?
}
public static void main(String[] args) {
System.out.println(fact(5));
// prints "5 * 4 * 3 * 2 * 1"
}
}
A simplified version of your code:
public int fact(int n)
{
if(n == 1){
return 1;
}
return n * (fact(n-1));
}
could be just:
public int fact(int n)
{
return n == 1 ? 1 : n * fact(n - 1);
}
but your code is not wrong, this is just another style (if you are not used to ternary operator keep the way it is). I prefer use the ternary operator in these cases (observe that the code is side effect free).
Works fine. You're not assigning it to anything. Here's a test that'll prove it works.
#Test
public void testYourFactorialMethod() {
assertEquals(120, fact(5));
}
public class Recursive {
public static void main(String[] argss) {
System.out.print(fac(3));
}
public static int fac(int n) {
int value = 0;
if (n == 0) {
value = 1;
} else {
value = n * fac(n - 1);
}
return value;
}
}
// out put 6
Try something like this:
(Or maybe try this directly)
public class factorial {
private static int factorial( int n ){
if (n > 1) {
return n * (factorial(n-1));
} else {
return 1;
}
}
public static void main(String[] args) {
System.out.println(factorial(100));
}
}
static int factorial(int x) {
int result;
if (x == 1) {
return 1;
}
// Call the same method with argument x-1
result = factorial(x – 1) * x;
return result;
}
For complete example check this
http://answersz.com/factorial-program-in-java-using-recursion/
It is totaly wrong to write Fibonacci with recursive methods!!
It is an old famous example for how a good/bad Algorythm affect any project
if you write Fibonatcci recursive, for calculating 120 you need 36 year toget the result!!!!!!
public static int Fibonacci(int x)
{ // bad fibonacci recursive code
if (x <= 1)
return 1;
return Fibonacci(x - 1) + Fibonacci(x - 2);
}
in dot net 4.0 there is a new type name BigInteger and you can use it to make a better function
using System;
using System.Collections.Generic;
using System.Numerics; //needs a ref. to this assembly
namespace Fibonaci
{
public class CFibonacci
{
public static int Fibonacci(int x)
{
if (x <= 1)
return 1;
return Fibonacci(x - 1) + Fibonacci(x - 2);
}
public static IEnumerable<BigInteger> BigFib(Int64 toNumber)
{
BigInteger previous = 0;
BigInteger current = 1;
for (Int64 y = 1; y <= toNumber; y++)
{
var auxiliar = current;
current += previous;
previous = auxiliar;
yield return current;
}
}
}
}
and you can use it like
using System;
using System.Linq;
namespace Fibonaci
{
class Program
{
static void Main()
{
foreach (var i in CFibonacci.BigFib(10))
{
Console.WriteLine("{0}", i);
}
var num = 12000;
var fib = CFibonacci.BigFib(num).Last();
Console.WriteLine("fib({0})={1}", num, fib);
Console.WriteLine("Press a key...");
Console.ReadKey();
}
}
}
and in this case you can calculate 12000 less than a second. so
Using Recursive methos is not always a good idea
Above code imported from Vahid Nasiri blog whiche wrote in Persian