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.
Related
I am trying to make a converter that converts decimal into binary, there is a catch tho, I can't use any other loops or statements except
while (){}
And I can't figure out how to start subtracting the number that fits into the decimal when it can and not using any if statements. Does anyone have any suggestions?
import java.util.Scanner;
public class Converter{
static Scanner input = new Scanner (System.in);
public static void main (String[] args){
System.out.println ("What is the number in the decimal system that you want to convert to binary?");
int dec = input.nextInt();
int sqr = 1024;
int rem;
while (dec != 0){
rem = dec / sqr;
sqr = sqr / 2;
System.out.print(rem);
}
}
}
Try this:
import java.util.Scanner;
public class Converter {
public static void main(String[] args) {
final Scanner input = new Scanner(System.in);
System.out.println("What is the number in the decimal system that you want to convert to binary?");
int dec = input.nextInt();
int div = 128;
while (div > 0) {
System.out.print(dec / div);
dec = dec % div;
div >>= 1; // equivalent to div /= 2
}
System.out.println();
}
}
Now, let's go through the code and try to understand what's going on. I'm assuming that the maximum size is 8 bits, so the variable div is set to 2n-1 where n = 1. If you need 16 bits, div would be 32768.
The programme starts from that value and attempts to do an integer division of the given number by the divider. And the nice thing about it is that it will yield 1 if the number is greater than or equal to the divider, and 0 otherwise.
So, if the number we're trying to convert is 42, then dividing it by 128 yields 0, so we know that the first digit of our binary number is 0.
After that, we set the number to be the remainder of the integer division, and we divide the divider by two. I'm doing this with a bit shift right (div >>= 1), but you could also use a divider-assignment (div /= 2).
By now, the divider is 64, and the number is still 42. If we do the operation again, we again get 0.
At the third iteration, we divide 42 by 32, and this yields 1. So our binary digits so far are 001. We set the number to be the remainder of the division, which is 10.
Continuing this, we end up with the binary number 00101010. The loop ends when the divider div is zero and there's nothing left to divide.
Try to understand, step by step, how the programme works. It's simple, but it can be very difficult to come up with a simple solution. In this case, it's applied mathematics, and knowing how integer maths work in Java. That comes with experience, which you'll get in due time.
Your code has some Problem. It is more easier to convert a decimal to binary. fro example:
int num = 5;
StringBuilder bin = new StringBuilder();
while (num > 0) {
bin.append(num % 2);
num /= 2;
}
System.out.println(bin.reverse());
I use StringBuilder to reverse my String and I prefer String because length of binary can be anything. if you use int or long, maybe overflow happen.
Update
if you you want to use primitive types only, you can do something like this but overflow may happen:
long reversedBin = 0, Bin = 0;
while (n > 0) {
reversedBin = reversedBin * 10 + (n % 2);
n /= 2;
}
while (reversedBin > 0) {
Bin = Bin * 10 + (reversedBin % 10);
reversedBin /= 10;
}
System.out.println(Bin);
Remember the algorithm to convert from decimal to binary.
Let n be a number in decimal representation:
digit_list = new empty stack
while n>0 do
digit = n%2
push digit in stack
n = n/2
end while
binary = new empty string
while digit_list is not empty do
character = pop from stack
append character to binary
end while
Java provides a generic class Stack that you can use as a data structure. You could also use lists, but remember to take the digits in the inverse order you have calculated them.
find the base 2 log of the number and floor it to find the number of bits needed. then integer divide by that bits place in 2's power and subtract that from the original number repeat until 0. doesn't work for negative. there are better solutions but this one is mine
int bits = (int) Math.floor(Math.log((double) dec) / Math.log((double) 2));
System.out.println("BITS:" + bits);
while (dec > 0) {
int twoPow = (int) Math.pow((double) 2, (double) bits);
rem = dec / twoPow;
dec = dec - rem * twoPow;
bits--;
System.out.print(rem);
}
This question already has answers here:
Java: parse int value from a char
(9 answers)
Closed 5 years ago.
I am trying to fetch second digit from a long variable.
long mi = 110000000;
int firstDigit = 0;
String numStr = Long.toString(mi);
for (int i = 0; i < numStr.length(); i++) {
System.out.println("" + i + " " + numStr.charAt(i));
firstDigit = numStr.charAt(1);
}
When I am printing firstDigit = numStr.charAt(1) on console. I am getting 1 which is expected but when the loop finishes firstDigit has 49.
Little confused why.
Because 49 is the ASCII value of char '1'.
So you should not assign a char to int directly.
And you don't need a loop here which keeps ovveriding the current value with charAt(1) anyway.
int number = numStr.charAt(1) - '0'; // substracting ASCII start value
The above statement internally works like 49 -48 and gives you 1.
If you feel like that is confusious, as others stated use Character.getNumericValue();
Or, although I don't like ""+ hack, below should work
int secondDigit = Integer.parseInt("" + String.valueOf(mi).charAt(1));
You got confused because 49 is ASCII value of integer 1. So you may parse character to integer then you can see integer value.
Integer.parseInt(String.valueOf(mi).charAt(1)):
You're probably looking for Character.getNumericValue(...) i.e.
firstDigit = Character.getNumericValue(numStr.charAt(1));
Otherwise, as the variable firstDigit is of type int that means you're assigning the ASCII representation of the character '1' which is 49 rather than the integer at the specified index.
Also, note that since you're interested in only a particular digit there is no need to put the statement firstDigit = numStr.charAt(1); inside the loop.
rather, just do the following outside the loop.
int number = Character.getNumericValue(numStr.charAt(1));
you only need define firstDigit as a char type variable, so will print as character.
since you define as int variable, it's value is the ASCII value of char '1': 49. this is why you get 49 instead of 1.
the answer Integer.parseInt(String.valueOf(mi).charAt(1)+""); is correct.
However, if we want to consider performace in our program, we need some improvements.
We have to time consuming methods, Integer.parseInt() and String.valueOf(). And always a custom methods is much faster than Integer.parseInt() and String.valueOf(). see simple benchmarks.
So, high performance solution can be like below:
int y=0;
while (mi>10)
{
y=(int) (mi%10);
mi=mi/10;
}
System.out.println("Answer is: " + y);
to test it:
long mi=4642345432634278834L;
int y=0;
long start = System.nanoTime();
//first solution
//y=Integer.parseInt(String.valueOf(mi).charAt(1)+"");
//seconf solution
while (mi>10)
{
y=(int) (mi%10);
mi=mi/10;
}
long finish = System.nanoTime();
long d = finish - start;
System.out.println("Answer is: " + y + " , Used time: " + d);
//about 821 to 1232 for while in 10 runs
//about 61225 to 76687 for parseInt in 10 runs
Doing string manipulation to work with numbers is almost always the wrong approach.
To get the second digit use the following;
int digitnum = 2;
int length = (int)Math.log10(mi));
int digit = (int)((mi/Math.pow(base,length-digitnum+1))%base);
If you want a different digit than the second change digitnum.
To avoid uncertainty with regards to floating point numbers you can use a integer math library like guavas IntMath
Let's take a look
System.out.println(numStr.charAt(1));
firstDigit = numStr.charAt(1);
System.out.println(firstDigit);
The result wouldn't be the same you will get
1
49
This happens because your firstDigit is int. Change it to char and you will get expected result
You can also do like below,
firstDigit = Integer.parseInt( numStr.charAt(1)+"");
So it will print second digit from long number.
Some things which have not been mentioned yet:
The second digit for integer datatypes is undefined if the long number is 0-9 (No, it is not zero. Integers do not have decimal places, this is only correct for floating-point numbers. Even then you must return undefined for NaN or an infinity value). In this case you should return a sentinel like e.g. -1 to indicate that there is no second digit.
Using log10 to get specific digits looks elegant, but they are 1. one of the numerically most expensive functions and 2. do often give incorrect results in edge cases. I will give some counterexamples later.
Performance could be improved further:
public static int getSecondDigit(long value) {
long tmp = value >= 0 ? value : -value;
if (tmp < 10) {
return -1;
}
long bigNumber = 1000000000000000000L;
boolean isBig = value >= bigNumber;
long decrement = isBig ? 100000000000000000L : 1;
long firstDigit = isBig ? bigNumber : 10;
int result = 0;
if (!isBig) {
long test = 100;
while (true) {
if (test > value) {
break;
}
decrement = firstDigit;
firstDigit = test;
test *= 10;
}
}
// Remove first
while (tmp >= firstDigit) {
tmp -= firstDigit;
}
// Count second
while (tmp >= decrement) {
tmp -= decrement;
result++;
}
return result;
}
Comparison:
1 000 000 random longs
String.valueOf()/Character.getNumericValue(): 106 ms
Log/Pow by Taemyr: 151 ms
Div10 by #Gholamali-Irani: 45 ms
Routine above: 30 ms
This is not the end, it can be even faster by lookup tables
decrementing 1/2/4/8, 10/20/40/80 and avoid the use of multiplication.
try this to get second char of your long
mi.toString().charAt(1);
How to get ASCII code
int ascii = 'A';
int ascii = 'a';
So if you assign a character to an integer, the integer will be holding the ASCII value of that character. Here I explicitly gave the values, in your code you are calling a method that returns a character, that's why you are getting ASCII instead of digit.
This is my function in Java:
public static String convertFromDecimal(int number, int base)
{
String result = "";
/*
* This while loop will keep running until 'number' is not 0
*/
while(number != 0)
{
result = (number%base) + result; // Appending the remainder
number = number / base; // Dividing the number by the base so we can get the next remainder
}
// If the number is already 0, then the while loop will ignore it, so we will return "0"
if(result == "")
{
return "0";
}
return result;
}
It works fine for numbers that convert to numbers not beginning with 0, if the number is supposed to have a zero at the start, it will not record it, could anyone tell me why?
For example, if I print out
convertFromDecimal(13,2) it returns
1101
Which is correct, but if I print out
convertFromDecimal(461,2), I get
111001101
Where the actual answer is
0000000111001101
So it's the same as my answer without the leading zeroes, if anyone knows why I would appreciate the help, thank you.
EDIT My question is different because I don't want 16 digits, I want the binary number of the given decimal, a calculator like this can explain what I want.
I assume you are looking to format all your answers as shorts (16 bits).
In this case, simply check the length of your current string, and add on zeroes as needed.
int zeroesRemaining = 16 - result.length();
for (int i = 0; i < zeroesRemaining; i++) {
result = "0" + result;
}
Alternatively, if you want to do it faster, use a StringBuilder.
int zeroesRemaining = 16 - result.length();
StringBuilder tempBuilder = new StringBuilder(result);
for (int i = 0; i < zeroesRemaining; i++) {
tempBuilder.insert(0, 0); //inserts the integer 0 at position 0 of the stringbuilder
}
return tempBuilder.toString(); //converts to string format
There is also probably a formatter that could do this, but I don't know of such.
If you want to change the number of zeroes to be the closest integer primitive, just set zeroesRemaining to be the (least power of 2 that is greater than the number of bits) minus (the number of bits).
Since you want fixed lengths for your result, in groups of 8 bits, the easiest way is to append 0 to the front of your result until its length is a multiple of 8.
That is as simple as
wile (result.length() % 8 > 0)
{
result = "0" + result;
}
return result;
Output for converting a number in decimal into its 1s complement and then again converting the number into decimal does not come as expected.
MyApproach
I first converted the number from decimal to binary. Replaced all Os with 1 and vice versa and then converted the number into decimal.
Can anyone guide me? What I am doing wrong?
Code:
public static int complimentDecimal(int num) {
int p = 0;
String s1 = "";
// Convert Decimal to Binary
while (num > 0) {
p = num % 2;
s1 = p + s1;
num = num / 2;
}
System.out.println(s1);
// Replace the 0s with 1s and 1s with 0s
for (int j = 0; j < s1.length(); j++) {
if (s1.charAt(j) == 0) {
s1.replace(s1.charAt(j), '1');
} else {
s1.replace(s1.charAt(j), '0');
}
}
System.out.println(s1);
int decimal = 0;
int k = 0;
for (int m = s1.length() - 1; m >= 0; m--) {
decimal += (s1.charAt(m) * Math.pow(2, k));
k++;
}
return decimal;
}
First of all you need to define the amount of Bits your binary representation should have or an complement representation does not make sense.
If you convert 100 the binary is 1100100
complement is 0011011 which is 27
now convert 27. Binary is 11011, complement 00100 which is 4.
Now define yourself a Bit length of 8.
100 is 01100100, complement 10011011, is 155
155 is 10011011, complement 01100100, is 100
Works because every binary representation has a length of 8 bits. This is absolutly necessary for the whole complement thing to make any sense.
Consider that you now have a limit for numbers that are convertable.
11111111 which is 255.
Now that we talked about that I will correct your code
static int MAX_BITS = 8;
static int MAX_INT = (int)Math.pow(2, MAX_BITS) - 1;
public static int complimentDecimal(int num)
{
// check if number is to high for the bitmask
if(num > MAX_INT){
System.out.println("Number=" + num + " to high for MAX_BITS="+MAX_BITS);
return -1;
}
// Your conversion works!
int p=0;
String s1="";
//Convert Decimal to Binary
while(num>0)
{
p=num%2;
s1=p+s1;
num=num/2;
}
// fill starting zeros to match MAX_BITS length
while(s1.length() < MAX_BITS)
s1 = "0" + s1;
System.out.println(s1);
//Replace the 0s with 1s and 1s with 0s
// your approach on that is very wrong
StringBuilder sb = new StringBuilder();
for(int j=0;j<s1.length();j++){
if(s1.charAt(j)=='0') sb.append("1");
else if(s1.charAt(j)=='1') sb.append("0");
}
s1 = sb.toString();
/*
for(int j=0;j<s1.length();j++)
{
if(s1.charAt(j)==0)
{
s1.replace(s1.charAt(j),'1');
}
else
{
s1.replace(s1.charAt(j),'0');
}
}
*/
System.out.println(s1);
int decimal=0;
int k=0;
for(int m=s1.length()-1;m>=0;m--)
{
// you don't want the char code here but the int value of the char code
//decimal += (s1.charAt(m) * Math.pow(2, k));
decimal+=(Character.getNumericValue(s1.charAt(m))*Math.pow(2, k));
k++;
}
return decimal;
}
Additional Note: Don't get bigger then MAX_BITS = 31 or you need to work with long instead of int in your method.
First of all you have to assign the replaced String to the already defined variable that is,
s1.replace(s1.charAt(j),'1');
it should be
s1 = s1.replace(s1.charAt(j),'1');
and the next case is, when you are changing in that order it would change all the characters similar to matched case
refer Replace a character at a specific index in a string?
String.Replace(oldChar, newChar) method returns a new string resulting from replacing all occurrences of oldChar in given string with newChar. It does not perform change on the given string.
The problem (OK, one of the problems) is here:
if(s1.charAt(j)==0)
Characters in Java are actually integers, in the range 0 to 65535. Each of those numbers actually means the character corresponding to that number in the Unicode chart. The character '0' has the value 48, not 0. So when you've created a string of '0' and '1' characters, the characters will have the integer values 48 and 49. Naturally, when you compare this to the integer 0, you'll get false no matter what.
Try
if(s1.charAt(j)=='0')
(Note: OK, the other answer is right--replace does not work. Not only are you using it incorrectly, by not assigning the result, it's not the right method anyway, because s1.replace(s1.charAt(j),'1') replaces all '0' with '1' characters; it doesn't replace character j. If you specifically want to replace the j'th character in a String with something else, you'll need to use substring() and build a new string, not replace().)
A couple other things to note: (1) Integers are not "decimal" or "binary". When your method gets the num parameter, this is just a number, not a decimal number or a binary number. It's represented in your computer as a binary number (unless you're using something like a Burroughs 3500, but I think all of those died before Java was invented). But it really isn't considered decimal, binary, octal, hex, ternary, or whatever, until you do something that converts it to a String. (2) I know you said not to post alternative approaches, but you could replace the entire method with just one line: return ~num;. That complements all the bits. If you were thinking that you couldn't do this because num was a decimal number, see #1. (3) "Compliment" means to say something nice about somebody. If you're talking about flipping all the bits, the correct spelling is "complement".
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.