LeetCode Notes: Number of 1 Bits

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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 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?

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;
};

Reference

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