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.