LintCode 365 - Count 1 in Binary

Count 1 in Binary

Count how many 1 in binary representation of a 32-bit integer.

上面程序中有三种方法，都来自编程之美
第一种：

原数除以2后，数字将减少一个0

若余数是1则，减少一个1，记录1的个数

若余数是0则，减少一个0，记录0的个数

第二种：

利用位运算

0&1 =0

1&1 =1

这个32位数与0000 0001 与运算的值是1，记录1的个数

即：count += num & 0x01

num右移一位

num=num>>1

    public int countOnes1(int num){
        int count = 0;
        while(num!=0){
            if(num%2==1)
                count++;
            num=num/2;
        }
        return count;
    }
    public int countOnes2(int num){
        int count = 0;
        while(num!=0){
            count +=num&0x01;
            num = num>>1;
        }
        return count;
    }
    public int countOnes3(int num){
        int count = 0;
        while(num!=0){
            num = num & (num-1);
            count++;
        }
        return count;
    }

http://www.chenguanghe.com/lintcode-count-1-in-binary/
// no need to iterate all 32 bits

public int countOnes(int num) {

        int res = 0;

        for(int i = 0 ; i < 32; i++) {

            if((num & (1<<i)) != 0)

                res++;

        }

        return res;

    }

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