191. Number of 1 Bits

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本文介绍了一种高效的方法来计算一个给定的32位无符号整数在其二进制表示中1的个数，即汉明重量。通过两种不同的算法实现：逐位检查和清除最低位的1，后者的时间复杂度为O(m)，其中m为1的个数。

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.

3 ms

class Solution { 
public:
    int hammingWeight(uint32_t n) {
        int ans = 0;
        while(n) {
            if (n & 1) ans++;
            n >>= 1;
        }
        return ans;
   }
};

better;

"n &= n - 1" is used to delete the right "1" of n. For example:

if n = 5 (101), then n-1 = 100, so n & (n-1) = 100, the right "1" is deleted;
if n = 6,(110), then n-1 = 101, so n & (n-1) = 100, the right "1" is also deleted;
and so on...

So time complexity is O(m), and m is the count of 1's, also m is less than or equal to 32.

class Solution { 
public:
    int hammingWeight(uint32_t n) {
        int ans = 0;
        while(n) {
            n &= n-1;
            ++ans;
        }
        return ans;
   }
};