動機

雖然說是水題，但是在LC的Solution中，卻有一個有趣的解法

Problem

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

Note:

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

 

Example 1:

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

Example 2:

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

Example 3:

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

 

Constraints:

  • The input must be a binary string of length 32.

 

Follow up: If this function is called many times, how would you optimize it?

Sol1

一直除2看有幾次奇數

class Solution:
    def hammingWeight(self, n: int) -> int:
        ret = 0
        while n != 0:
            if n % 2 == 1:
                ret += 1
            n = n//2
        return ret

Case study

10轉成binary，是 01010

如果10-1會從離左手邊最近的1借位，所以會變成 01001

注意到，到左手邊最近的1之間的所有0，都變成1了!!

同時因為借位，所以最近的1變成0

最後只要與原本的數字做AND，就會把 左手邊最近的1之間的所有1 ，都變成0了!!

01010 (10)
[-1]
01001 (9)
[&n]
01000 (8)

這樣只要到0之前一直這樣做就好了!!

public int hammingWeight(int n) {
    int sum = 0;
    while (n != 0) {
        sum++;
        n &= (n - 1);
    }
    return sum;
}