The number of 1 in binary _java

Preface


will be writing some regular interview questions recently, the test will share with you after the





Displacement Method


is only suitable for positive numbers:




The
shift method is to determine whether the minimum bit of n is 1, and the time complexity is O (logn)

Copy Code code as follows:

int nativeonenum (int n)
{
int count = 0;

while (n) {
if (n & 1) count + +;
n >>= 1;
}

return count;
}

is no problem for positive numbers, but if n is negative, here's the problem, with negative 8 as an example, the binary complement form is 11111111|11111111|11111111|11111000|, and the right one becomes the 11111111|11111111 |11111111|11111100|, because it is a negative number, so the sign bit will continue to fill 1, resulting in the last 1, the death cycle





in this case, we can use the variable flag = 1, from right to left and n comparison, 32-bit int up to 32 times to know the number of n 1

Copy Code code as follows:

int onenum (int n)
{
int count, flag;

for (count = 0, flag = 1; flag; flag <<= 1) {
if (flag & N) count + +;
}

return count;
}

Rapid Method


This solution is that the number of binary 1 is only related to 1 of the number of digits, N &amp; (n-1) quickly remove the leftmost 1, such as 7 (0111) &amp; 6 (0110) = 6 (0110), quickly remove the leftmost 1

Copy Code code as follows:

int quickone (int n)   
{  
    I NT count = 0;  

    while (n) {  
       ;  Count ++;  
        n = n & (n-1);  
&nbs p;  }  

    return count;  
}