[leetcode]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?

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题目：除了一个元素以外其它元素出现3次，而只有一个元素出现一次

思路：将每个元素转化成二进制并将其对应的bit位置1，所以出过三次的元素对应的位应该是3的倍数，而最后只会剩下那只出现一次元素所对应的二进制位。

int singleNumber(int a[], int n) {
int bitCount[32] = {0};

for(int i = 0; i < n; i++){
int bit = 1;
for(int j = 0; j < 32; j++){
if(bit & a[i]){
bitCount[j] = (bitCount[j] + 1) % 3;
}
bit <<= 1;
}
}

int retNum = 0;

for(int i = 0; i < 32; i++){

retNum = retNum * 2 + bitCount[31 - i];
}

return retNum;
}

如果对其进一步可作扩展，除一个元素出现2次其它都出现3次，也可以用此方法。

如一个元素出现两次，一个出现1次，其它出现4次。也可以