Single number II

/*Test instructions: Give an array of integers, each element appears three times, only one element appears once, find out the number (target number) solution: Use bit operation, count the number of each bit 1 of the whole array, and then 3 modulo, because if the first bit of the target number is 0, then bit%3 is 0, no is 1, the bit is left-shifted, and the result is added directly. */classSolution { Public:    intSinglenumber (intA[],intN) {intBit,res =0;  for(inti =0; I < +; i++) {bit=0;  for(intj =0; J < N; J + +) {bit+ = (a[j]>>i) &1; } Res|= (bit%3) <<i; }        returnRes; }};

Single number II