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.
1 public class Solution { 2 // you need to treat n as an unsigned value 3 public int hammingWeight(int n) { 4 n = n - ((n >>> 1) & 0x55555555); 5 n = (n & 0x33333333) + ((n >>> 2) & 0x33333333); 6 n = (n + (n >>> 4)) & 0x0f0f0f0f; 7 n = n + (n >>> 8); 8 n = n + (n >>> 16); 9 return n & 0x3f; 10 } 11 }
1 public class Solution { 2 // you need to treat n as an unsigned value 3 public int hammingWeight(int n) { 4 int result = 0; 5 while(n > 1){ 6 result += n % 2; 7 n = (int)(n / 2); 8 } 9 if(n == 1) result ++; 10 return result; 11 } 12 }
it fail at n == 2147483648：
Because 2147483648 is not greater than 2147483648, and then shift to left 1 bit make bit == 0 instead of 2147483648 * 2. Then it fall into the infinity loop where bit always be 0.
1 public class Solution { 2 // you need to treat n as an unsigned value 3 public int hammingWeight(int n) { 4 int result = 0; 5 for(int i = 0; i < 32; i ++){ 6 if(((n>>i)&1) == 1) result ++; 7 } 8 return result; 9 } 10 }