191. Number of 1 Bits

191. 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). For example, the 32-bit integer ’11’ has binary representation 00000000000000000000000000001011, so the function should return 3.

Credits: Special thanks to @ts for adding this problem and creating all test cases.

public class Solution { // you need to treat n as an unsigned value public int hammingWeight(int n) { int cnt = 0; // loop times = number of 1's in the n while(n != 0) { n = n & (n - 1); cnt ++; } return cnt; }}

public class Solution { // you need to treat n as an unsigned value public int hammingWeight(int n) { int bits = 0; int mask = 1; for (int i = 0; i < 32; i++) { if ((mask & n) != 0) bits++; mask = mask << 1; } return bits; }}

/** * @param {number} n - a positive integer * @return {number} */var hammingWeight = function(n) { return n.toString(2) !== '0' ? (n.toString(2).match(/1/g)).length : 0;};

/** * @param {number} n - a positive integer * @return {number} */var hammingWeight = function(n) { return n.toString(2).split('').filter((ele) => ele === '1').length};

public class Solution { // you need to treat n as an unsigned value public int hammingWeight(int n) { int mask = 0x00000001; int count = 0; while(n != 0) { if((n & mask) == 1) { count++; } n >>>=1; } return count; }}

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