Problem
Write a function that takes an unsigned integer and returns the number of ’1′ bits it has (also known as the Hamming weight).
For example, the 32-bit integer ’11′ has binary representation 00000000000000000000000000001011, so the function should return 3.
Java Solution
public int hammingWeight(int n) { int count = 0; for(int i=1; i<33; i++){ if(getBit(n, i) == true){ count++; } } return count; } public boolean getBit(int n, int i){ return (n & (1 << i)) != 0; } |
For java, you can just convert the value to String and remove all 0s, and then return the length of that String.
There’s no-loop solution (C code, but can be easily adopted to Java):
int pop(unsigned x)
{
x = x - ((x >> 1) & 0x55555555);
x = (x & 0x33333333) + ((x >> 2) & 0x33333333);
x = (x + (x >> 4)) & 0x0F0F0F0F;
x = x + (x >> 8);
x = x + (x >> 16);
return x & 0x0000003F;
}
Got from here.
public int count(int num) {
int count = 0;
while(num > 0) {
count += num & 1;
num >>= 1;
}
return count;
}
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The for loop needs to run from i=0 to t<32
did you tried it ? n = 5
int nuum= 0;
while (n != 0)
{
int x = n & 1;
if (x == 1) nuum++;
n >>= 1;
}
return nuum;
int hammingWeight(int n) {
int num = 0;
while (n != 0) {
n = n & (n-1);
num++;
}
return num;
}
We could not use bit operators, just keep on dividing and modding by 2 until n==0; every time n%2==1, just add 1 to int answer.
Loop should run from 0 to 31
Why don’t you use this :
public class Solution {
public int hammingWeight(int n) {
return Integer.bitCount(n);
}
}