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 = 0;    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=32;    int ret = 0;    while(m--)    {        ret = ret >> 1;        ret &= 0x7fffffff;        int sum = 0;        for(int i=0;i<n;i++)        {            sum += A[i] & 0x00000001;            A[i] = A[i] >> 1;        }        if(sum%3){            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=0, two=0, thr=0;    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;    }

Hope this may help.

Single Number