LeetCode-137 Single number II

Given an array of integers, every element appears three times except for one. Find the single one.

Note:
Your algorithm should has a linear runtime complexity. Could you implement it without using extra memory?

Ideas:

Consider all binary representations, if we take the sum of all the numbers on the first th position and the 3, then there will only be two results 0 or 1 (according to test instructions, 3 0 or 3 1 Add the remainder to 0). So the result of the remainder is that "single number".

A straightforward implementation is to use an array of size 32 to record the and on all bits.

intSinglenumber (intA[],intN) {intCOUNT[32] = {0}; intresult = 0;  for(inti = 0; I < 32; i++) {     for(intj = 0; J < N; J + +) {      if((A[j] >> i) & 1) {Count[i]++; }} result|= ((Count[i]% 3) <<i); }  returnresult;}

Improved:

This algorithm has improved space and can be used with mask variables:

1. onesMask variable that represents only one occurrence of the first th bit
2. twosMask variable representing only two occurrences of the first th bit

Assuming that there are 3 consecutive 5 occurrences at the beginning of the array, the change is as follows:

ones = 101twos = 0--------------ones = 0twos = 101--------------ones = 0twos = 0--------------  
When the first th bit appears 3 times, we set the ones first twos th bit to 0. The final answer is ones.

The code is as follows:

 Public intSinglenumber (int[] A) {intone = 0; intboth = 0; inttmp;  for(inti=0; i<a.length; i++) {= (A[i] & one) |both ; One= one ^A[i]; TMP= ~ (One &); One= one &tmp; both= &tmp; }        returnOne ; }

The above analysis section references the self-propelled cool: http://www.tuicool.com/articles/fAZZv2a

LeetCode-137 Single number II