Two numbers (three numbers) that occur once in an array & ask the last bit to be 1

For two numbers, for the result, the remaining bit1 are different or differentiated.

The following solution is very concise:

int lastBitOf1(intNumber ) {    returnNumber & ~ (number-1);} voidGettwounique (vector<int>::iterator Begin, vector<int>::iterator End, vector<int>&unique) {    intXorresult =0;  for(vector<int>::iterator iter = begin; Iter! = END; ++iter) Xorresult^= *ITER; intdiff =lastBitOf1 (Xorresult); intFirst =0; intSecond =0;  for(vector<int>::iterator iter = begin; Iter! = END; ++ITER) {        if(diff & *iter) first^= *ITER; ElseSecond^= *ITER;    } unique.push_back (first); Unique.push_back (second);}

For cases with three numbers, it's a bit more complicated:

To use a different or result with all the different or the last one, then different or. As follows:

http://blog.csdn.net/sunmenggmail/article/details/8035008

1. for (iter = Numbers.begin (); ITER! = Numbers.end (); ++iter)
2. Flags ^= lastBitOf1 (xorresult ^ *iter);
3. Flags = LASTBITOF1 (flags);

Two numbers (three numbers) that occur once in an array & ask the last bit to be 1