计算二进制中1的个数

int countones1(unsigned int n){    int ones = 0;    while(0 < n)    {       ones += (n & 1);       n >> 1;    }    return ones;}

复杂度O(logn)

int countones2(unsigned int n){    int ones = 0;    while(0 < n)    {       ones++;       n &= n-1;    }    return ones;}

复杂度O(countOnes(n))

#define POW(c) (1 << (c))#define MASK(c) (((unsigned int)-1) / (POW(POW(c)) + 1)) #define ROUND(n, c) (((n) & MASK(c)) + (((n) >> POW(c)) & MASK(c)))int countOnes3(unsigned int n){    n = ROUND(n, 0);    n = ROUND(n, 1);    n = ROUND(n, 2);    n = ROUND(n, 3);    n = ROUND(n, 4);    return n;}

复杂度为O(log2(W)),W=log2(n)为整数的位宽,即为O（loglogn）