【Leetcode】Single Number II

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?

class Solution {
public:
    int singleNumber(int A[], int n) {
        int * record = new int[32];
        int ret = 0;
        
        for (int i = 0; i < 32; ++i) {
            record[i] = 0;
            for (int j = 0; j < n; ++j) {
                if ((A[j] >> i) & 1 == 1) {
                    ++record[i];
                }
            }
            if (record[i] % 3 == 1) {
                ret |= (1 << i);
            }
        }
        return ret;
    }
};

与上一题Single Number相似的思路，用位运算来实现，不过不能直接用二进制的异或，而是记录每个bit上1出现的次数。所有出现过3次的数，各个为1的位也必定重复出现了3次。所以记录每个位上1出现的总次数，如果某位上1出现的次数不是3的倍数，那么该位上多出来的1是那个single number带来的。由此可以还原出所求的数字。无论该数出现1次还是2次或者4次，都可以找出来。