137. Single Number II

Given an array of integers, every element appears three times except for one, which appears exactly once. Find that single one.

Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

含义：数组中每个整数都出现了三次，只有一个出现了一次，找出来

    public int singleNumber(int[] nums) {
//        我们可以建立一个32位的数字，来统计每一位上1出现的个数，我们知道如果某一位上为1的话，那么如果该整数出现了三次，对3去余为0，
// 我们把每个数的对应位都加起来对3取余，最终剩下来的那个数就是单独的数字
        int res = 0;
        for (int i = 0; i < 32; ++i) {
            int sum = 0;
            for (int j = 0; j < nums.length; ++j) {
                sum += (nums[j] >> i) & 1;
            }
            res |= (sum % 3) << i;
        }
        return res;        
    }