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How to calculate number of 1s in a binary number?
9 Réponses
+ 2
public class IntegerToBinary {
public static void main(String[] args) {
int x = 102;
int cnt=0;
String y = Integer.toBinaryString(x);
System.out.println("Binary conversion is: " + y);
for (int i=0; i<y.length();i++)
{
if (y.substring(i, i+1).equals("1"))
{
cnt =cnt+1;
}
}
System.out.println("1's are: " + cnt);
//one more way
cnt=0;
char[] z = y.toCharArray();
for (int i=0; i<z.length;i++)
{
if (Character.toString(z[i]).equals("0"))
{
cnt =cnt+1;
}
}
System.out.println("0's are : "+cnt);
}
}
+ 1
haha I mean in assembly language
+ 1
how about in java. I just want to know how it works.
+ 1
public class solution{
public static void main(String arcs[]){
string binnumber = Integer.toString(100,2);
//converting 100 into binary
int count = 0;
for(int i=0; i<binnumber.length(); i++){
String c = Character.toString(text.charAt(i))
if(c.equals("1"))
count++;
}
System.out .println(count);
}
}
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use a calculator
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1s ?
Um....Can you tell me the thing you confused :)?(Becus some word you use I'm falling in confused too)
If I can give advice I will
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I am a bit confused.. For example, x=5, print number of 1s =2
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Ok I think he help you for now
0
yea he is