【Leetcode长征系列】Single Number II

原题：

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： 用一个32位的数组存每一位bit值之后。得到答案后每一位除三取余即为答案。

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;
    }
};

我本觉得

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

不过全部报错

【Leetcode长征系列】Single Number II