191. Number of 1 Bits
Problem
Tags: Bit Manipulation
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, 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?
Code
C
// 191. Number of 1 Bits (5/8/54369)
// Runtime: 3 ms (40.72%) Memory: 5.43 MB (38.83%)
int hammingWeight (uint32_t n) {
uint8_t count = 0;
while (n) {
n = n & (n - 1);
count++;
}
return count;
}
GO
// 191. Number of 1 Bits (5/9/54369)
// Runtime: 0 ms (90.80%) Memory: 1.84 MB (43.77%)
func hammingWeight(num uint32) int {
count := 0
for num > 0 {
num = num & (num - 1)
count++
}
return count
}
JS
// 191. Number of 1 Bits (3/10/53745)
// Runtime: 80 ms (57.00%) Memory: 40.46 MB (94.65%)
/**
* @param {number} n - a positive integer
* @return {number}
*/
function hammingWeight(n) {
return [...n.toString(2)].filter((x) => x === "1").length;
}
TS
// 191. Number of 1 Bits (5/20/54369)
// Runtime: 109 ms (33.99%) Memory: 44.26 MB (87.93%)
function hammingWeight(n: number): number {
let count = 0;
while (n) {
count++;
n = n & (n - 1);
}
return count;
};