[LeetCode] 191. Number of 1 Bits

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.

Follow up:

If this function is called many times, how would you optimize it?
这道题是count一共有多少位是1，想法很简单。还是用最右边为1其余为0的模，与原值异或。这样可以依次从右往左把1改变为0。每一次这个操作都计数一次。一直到所有位置都为0就停止。时间复杂度O(N)，空间复杂度为O(1)。

class Solution:
    def hammingWeight(self, n: int) -> int:
        count = 0
        while n != 0:
            right_one = n & ((~n) + 1)
            count += 1
            n ^= right_one
        return count