【Leetcode長征系列】Single Number II

原題：

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?

思路： 用一個32位的數組存每一位bit值之後。得到答案後每一位除三取餘即爲答案。

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

我本覺得

res |= (bitnum[i]%3)<<i;

可以用

res =res+ (bitnum[i]%3)<<i;

代替，

或者是

bitnum[i] += (A[j]>>i)&1;

可以用

bitnum[i] = bitnum[i]+ (A[j]>>i)&1;

代替

不過全部報錯