【Leetcode】 - Single Number II

Problem Discription:

Suppose the array A has n items in which all of the numbers apear 3 times except one. Find the single number.

 int singleNumber2(int A[], int n) {
    int ret = ;
    while(n--)
    {
        ret ^= A[n];
    }
    return ret;
}

Related Problem:

Suppose the array A has n items in which all of the numbers apear twice except one. Find the single number.

Solution 1:

Suppose the required return value is ret. Each bit of ret is calculated by the respective bit of A[0:n-1].

    int singleNumber3(int A[], int n) {
    int m=;
    int ret = ;
    while(m--)
    {
        ret = ret >> ;
        ret &= 0x7fffffff;
        int sum = ;
        for(int i=;i<n;i++)
        {
            sum += A[i] & 0x00000001;
            A[i] = A[i] >> ;
        }
        if(sum%){
            ret |= 0x80000000;
        }
    }
    return ret;
}

Solution2: Solution1 needs 32*n passes of loop. Solution2 is found on the Internet, see http://blog.csdn.net/bigapplestar/article/details/12275381 . However the bit operation is confused. Therefore I write solution3, and try to explain it.

    public int singleNumber(int[] A) {
    int once = 0;
    int twice = 0;
    for (int i = 0; i < A.length; i++) {
        twice |= once & A[i];
        once ^= A[i];
        int not_three = ~(once & twice);
        once = not_three & once;
        twice = not_three & twice;
    }
    return once;
} 

Solution3:

My solution3 seems more easy-to-understand compared with solution2. Suppose the i-th bit of one, two, thr (onei, twoi, thri) is used to count how many bits in A[0-n-1]i is 1.

# onei twoi thr_i

1 1       0      0

2 0       1      0

3 0       0      0

4 1       0      0

5 0       1      0

6 0       0      0 ....

so we have:

if(A[i] == 1)

　　if(one == 0) one = 1;

　　else one = 0;

　　　　if(two == 0) two = 1;

　　　　else two = 0;

　　　　　　if(thr == 0) thr = 1;

　　　　　　else thr = 0;

　　when thr=1, one=two=0;

So with the bit operation we have an easy-to-understand version as below:

int singleNumber3(int A[], int n) {
    int one=, two=, thr=;
    while(n--)
    {
        //thr ^= (one & two & A[n] );
        two ^= one & A[n];
        one ^= A[n];
        thr = one & two;
        one = (~thr) & one;
        two = (~thr) & two;
        //thr = (~thr) & thr;
    }
　　 return one;
}

Hope this may help.