[Leetcode 137, Medium] Single Number II

Problem:

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?

Analysis:

解決方法是按（2進制或bit）位展開每一個數並相加。因爲是除了一個元素以爲，所有元素的出現都是3個。相加完畢，在例外元素bit值爲1的那一位必定是3×k+1的形式，bit爲0的那一位是3×k的形式。（k是某一個正整數）所以，模3以後剩下的值，再變成10進制，就是所要找的元素。

需要注意的是：按bit展開不需要預先做好。使用(A[i] >> j) & 1可以得到某個指定的展開位的值。

衍生問題：(a) every element appears k (k > 1) times except for one (b) every element appears a multiple of three times except for one, and the exceptional element appears once.

Solution:

C++:

int singleNumber(int A[], int n) 
{  
    int bitArray[32] = {0};  
      
    int rval = 0;  
    for(int j = 0; j < 32; ++j) {  
        for(int i = 0; i < n; ++i)  
            bitArray[j] += (A[i] >> j) & 1;  
              
        rval |= ((bitArray[j] % 3) << j);  
    }  
      
    return rval;  
}