LEETCODE: Single Number II

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本文介绍了一种算法，用于在一个整数数组中找出仅出现一次的元素，所有其他元素都出现了三次。该算法使用了一个32个元素的容器来记录各个位上数字出现的次数，并最终通过位运算返回唯一的整数。

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?

用一个32个元素的容器记录各个位上数字出现的记录！

class Solution {
public:
    int singleNumber(int A[], int n) {
        vector<int> bits(32);
        int result = 0;
        for(int ii = 0; ii < n; ii ++) {
            // Check every bit of A[ii]
            int base = 1;
            for(int jj = 0; jj < 32; jj ++) {
                int temp = A[ii] & base;
                if(temp != 0) {
                    bits[jj] ++;
                }
                base = base << 1;
            }
        }
        
        
        for(int ii = 0; ii < 32; ii ++) {
            if(bits[ii] % 3 == 1)
                result |=  1 << ii;
        }
        
        return result;
    }
};