Leetcode[191] Number of 1 Bits

###Task1 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.

###Python

class Solution(object):
    def hammingWeight(self, n):
        """
        :type n: int
        :rtype: int
        """
        ret = 0
        for i in range(32):
            cur = (n >> i) & 1
            if cur == 1:
                ret += 1
        return ret
            

###Java

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

###Points

• (n » a) & 1 –> get one bit
• n ^ (1 « a) –> rotate one bit

1. 输入值n可能为负数（但应视其为无符号整数，但Java中实际上是没有无符号整数的）

1. 使用无符号右移操作，可以忽略符号位。