[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?

 

方法一：创建一个长度为 多少个bit/sizeof(int) * 8的数组 count[sizeof(int)*8]，count[i] 表示在在 i 位
出现的 1 的次数。如果 count[i] 是 3 的整数倍，则忽略；否则就把该位取出来组成答案。

 

class Solution
{
    public:
        int singleNumber(int A[], int n)
        {
            vector<int> cnt;
            int width = sizeof(int)*8;
            cnt.resize(width, 0);
            int res = 0;

            for(int i  = 0; i< n; i++)
            {
                for(int j  = 0; j< width; j++)
                {
                    cnt[j] += (A[i] >> j) & 0x1;
                    //cnt[j] %= 3;
                }

            }
            for(int j  = 0; j< width; j++)
            {
                //cout << "j :\t" <<j <<"\t cnt\t"<< cnt[j]<<endl;
                if(cnt[j]%3 != 0)
                    res ^=  ( 1 << j);
            }
            return res;
        }
} ;