When writing leetcode, I have encountered several problems that need to quickly find the number of 1 in a binary sequence, so here is a summary of several methods.
Bitwise judgment
This is the simplest and most intuitive way to determine whether each digit of an integer is 1 in turn.
Judge the first bit: n & 1
Judge the second bit: n & (1 << 1)
Judge the nth bit: n & (1<<(n-1))
class Solution {
public:
int hammingWeight(uint32_t n) {
int ret = 0;
for (int i = 0; i < 32; i ++) {
if (n & (1 << i)) {
ret ++;
}
}
return ret;
}
};
Shortcut method: N & (N-1)
Using this method is based on the fact:
一个数 n 与一个比它小 1 的数（n−1）进行与运算（&）之后，得到的结果会消除 n 中最低位的 1。
For example:
7&6
00111 & 00110 =00110
It can be seen that the result of n&(n−1) is the value after replacing the lowest bit of n with 0.
This algorithm is used to eliminate 1 each time, and the counter is +1 each time until n is 0.
class Solution {
public:
int hammingWeight(uint32_t n) {
int res = 0;
while (n) {
n = n&(n - 1);
res++;
}
return res;
}
};