191.[Leetcode]Number of 1 Bits

//将转化好的二进制字符串存入s中，然后从s中判断有多少的1
// 从10进制到其他进制，都是对其求余数
class Solution {
public:

int hammingWeight(uint32_t n) {
string s = "";
char temp[1];
//循环求余数
int y(0);
while(n != 0){
y = n%2;
sprintf(temp,"%d",y);
s += temp;
n = n/2;
}
int count(0);
for(int i = 0; i < s.size(); i++)
if(s[i] == '1') count++;
return count;
}

};

int hammingWeight(uint32_t n) {
int cout(0);
while(n > 0) {
n = n&(n-1);
cout++;
}
return cout;
}