Number of 1 Bits
Write a function that takes an unsigned integer and returns the number of '1' bits it has (also known as the Hamming weight).
Note:
- Note that in some languages such as Java, there is no unsigned integer type. In this case, the input will be given as a signed integer type. It should not affect your implementation, as the integer's internal binary representation is the same, whether it is signed or unsigned.
-
In Java, the compiler represents the signed integers using 2's complement notation. Therefore, in Example 3 above, the input represents the signed integer.
-3.
Follow up: If this function is called many times, how would you optimize it?
Example 1:
Input: n = 00000000000000000000000000001011 Output: 3 Explanation: The input binary string 00000000000000000000000000001011 has a total of three '1' bits.
Example 2:
Input: n = 00000000000000000000000010000000 Output: 1 Explanation: The input binary string 00000000000000000000000010000000 has a total of one '1' bit.
Example 3:
Input: n = 11111111111111111111111111111101 Output: 31 Explanation: The input binary string 11111111111111111111111111111101 has a total of thirty one '1' bits.
Constraints:
-
The input must be a binary string of length
32
Solution:
public class Solution {
// you need to treat n as an unsigned value
public int hammingWeight(int n) {
return
Integer.toBinaryString(n).replaceAll("0","").length();
}
}
The above solution is one of the easiest solution using inbuild methods in
Java. First convert integer to binary string and using regex we are removing
all 0's from the string. Finally, returning the length of the string.
However, the above solution is more easy, but it consumes more run time. Thus
to avoid more runtime we are going to find another solution for solving this
problem. [Solution will be accepted in Leetcode].
Solution 2:
public class Solution {
// you need to treat n as an unsigned value
public int hammingWeight(int n) {
int count = 0;
for(int i=0;i<32;i++){
if(((n>>i)&1) ==1){
count++;
}
}
return count;
}
}
Here, in this solution, the basic idea is we are going to do right shift each
time(till end of the binary digits) and doing Bitwise AND operation, we are
going to find total count of 1's. The reason behind to use 32 in iteration is
given in problem description.
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