# LeetCode Notes: Number of 1 Bits

## Question

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 string00000000000000000000000000001011has a total of three '1' bits.

**Example 2:**

Input:n = 00000000000000000000000010000000

Output:1

Explanation:The input binary string00000000000000000000000010000000has a total of one '1' bit.

**Example 3:**

Input:n = 11111111111111111111111111111101

Output:31

Explanation:The input binary string11111111111111111111111111111101has 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?

## Solution One

**Analysis:**

Observe the characteristics of `n & 1`

n: 110010101001

1: 000000000001

n & 1: 000000000001

You can see that the result depends on the number in the same position as the number 1 in n, where the last number of n is 1, so the result is 1.

If every binary digit of a number uses `& 1`

, you can use this feature to count the number of ones. We use `1<<i`

to shift left by n places.

**Code:**

```
/**
* @param {number} n-a positive integer
* @return {number}
*/
var hammingWeight = function(n) {
let result = 0;
for(let i = 0; i <32; i++){
/**
n: 110010101001
1<<i: 000000000001
1 Shift i bits to the left, then each loop finds 1<<i the penultimate number 1 and the penultimate number of n (do not know whether it is 0 or 1), and perform the AND operation, and the result is either 0, or not 0, when it is not 0, it means 1
*/
if((n & (1 << i)) !== 0){
result ++;
}
}
return result;
};
```

## Solution Two

**Analysis:**

Observe `n = 6`

, `n.toString(2)`

to obtain binary numbers

n: 110

n-1: 101

n & (n-1): 100

`n & (n-1)`

will convert the lowest 1 of n to 0, and do so in a loop until the result is 0, and the number of 1s can be counted

**Code:**

```
/**
* @param {number} n-a positive integer
* @return {number}
*/
var hammingWeight = function(n) {
let result = 0;
while(n){
// n & (n-1) convert the lowest 1 of n to 0
n &= n-1;
result++;
}
return result;
};
```

## Comments