[Leetcode]191. Number of 1 Bits

191. Number of 1 Bits

• 本题难度: Easy
• Topic: Bit Manipulation

Description

Write a function that takes an unsigned integer and return the number of '1' bits it has (also known as the Hamming weight).

Example 1:

Input: 00000000000000000000000000001011
Output: 3
Explanation: The input binary string 00000000000000000000000000001011 has a total of three '1' bits.
Example 2:

Input: 00000000000000000000000010000000
Output: 1
Explanation: The input binary string 00000000000000000000000010000000 has a total of one '1' bit.
Example 3:

Input: 11111111111111111111111111111101
Output: 31
Explanation: The input binary string 11111111111111111111111111111101 has a total of thirty one '1' bits.

Note:

Note that in some languages such as Java, there is no unsigned integer type. In this case, the input will be given as signed integer type and should not affect your implementation, as the internal binary representation of the integer 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 above the input represents the signed integer -3.

我的代码

class Solution(object):
    def hammingWeight(self, n):
        """
        :type n: int
        :rtype: int
        """
        sn  =  str(bin(n))
        count = 0
        for i in sn:
            if i == '1':
                count += 1
        return count

别人的代码

1. Built in solution with built-in function:

Built in solution with built-in function:

2. Using bit operation to cancel a 1 in each round

def hammingWeight(self, n):
    """
    :type n: int
    :rtype: int
    """
    c = 0
    while n:
        n &= n - 1
        c += 1
    return c```
# 思路
我的思路还是从字符串的角度处理，而在别人代码的方法二中。每次n&(n-1)都可以消除一个1。
比如：
xxxx1 & xxxx0 = xxxx0
xx10..0(k个0) & xx0 1...1(k个1) = xx 0 0...0 (k+1个1) 
- 出错
注意一开始输入对n为十进制