[leetcode]Single Number II

此题就是对32个位分别数1的个数，如果是3的倍数，那么所求数该位为0，否则该位为1。

public class Solution {
    public int singleNumber(int[] A) {
        int ans = 0;
        for (int i = 0; i < 32; i++)
        {
            int cnt = 0;
            for (int j = 0; j < A.length; j++)
            {
                if ((A[j] & (1 << i)) != 0) cnt++;
            }
            if (cnt % 3 != 0) ans |= (1 << i);
        }
        return ans;
    }
}

第二刷：其实数组不需要，要注意位运算的优先级低

class Solution {
public:
    int singleNumber(int A[], int n) {
        int bits[32] = {};
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < 32; j++) {
                int k = (A[i] & (1 << j)) == 0 ? 0 : 1;
                bits[j] = (bits[j] + k) % 3;
            }
        }
        int r = 0;
        for (int i = 0; i < 32; i++) {
            if (bits[i] != 0) {
                r |= (1 << i);
            }
        }
        return r;
    }
};