Write a function that takes the binary representation of 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, the input represents the signed integer.
-3
.
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
.
Follow up: If this function is called many times, how would you optimize it?
Companies: Apple, Box, Amazon, Facebook, Qualcomm, Microsoft, Adobe, Cisco, Bloomberg, Google, Intel, Mindtickle
Related Topics:
Divide and Conquer, Bit Manipulation
Similar Questions:
- Reverse Bits (Easy)
- Power of Two (Easy)
- Counting Bits (Easy)
- Binary Watch (Easy)
- Hamming Distance (Easy)
- Binary Number with Alternating Bits (Easy)
- Prime Number of Set Bits in Binary Representation (Easy)
// OJ: https://leetcode.com/problems/number-of-1-bits/
// Author: github.com/lzl124631x
// Time: O(1)
// Space: O(1)
class Solution {
public:
int hammingWeight(uint32_t n) {
int ans = 0;
for (; n; n -= (n & -n)) ++ans;
return ans;
}
};
Or
// OJ: https://leetcode.com/problems/number-of-1-bits/
// Author: github.com/lzl124631x
// Time: O(1)
// Space: O(1)
class Solution {
public:
int hammingWeight(uint32_t n) {
int ans = 0;
for (; n; n = n & (n - 1)) ++ans;
return ans;
}
};
Or return __builtin_popcount(n);