Single Number II

Single Number II

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

Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?
取得每一位的加总和，然后 mod 3, 如果==0， 那么该位一定为0(表示那个数在该位为0)，否则该位为1 (那个数如果出现两次，那么取模后是2，如果出现一次，取模后是1)

public static int singleNumber(int[] A) {
int bit = 0;
for (int i = 0; i < 32; i++) {
int sum = 0;
for (int j = 0; j < A.length; j++) {
sum += (A[j] & (1 << i)) >> i;
}
bit |= (sum % 3 == 0 ? 0 : 1) << i; // in this case, the one can appear either once or twice.
}
return bit;
}