Gray Code

The gray code is a binary numeral system where two successive values differ in only one bit.

Given a non-negative integer n representing the total number of bits in the code, print the sequence of gray code. A gray code sequence must begin with 0.

For example, given n = 2, return [0,1,3,2]. Its gray code sequence is:

00 - 0
01 - 1
11 - 3
10 - 2

就是找规律，把上一层得到的结果逆序前面一位置1，然后把这些新数加回结果中

public class Solution {
    //Time: O(n^2)  Space: O(n^2)
    public ArrayList<Integer> grayCode(int n) {
        ArrayList<Integer> res = new ArrayList<Integer>();
        res.add(0);
        if (n <= 0) {
            return res;
        }
        
        res.add(1);
        for (int i = 1; i < n; i++) {
            for (int j = res.size() - 1; j >= 0; j--) {
                int cur = res.get(j);
                cur |= (1 << i);
                res.add(cur);
            }
        }
        return res;
    }
}

Single Number I & II

Given an array of integers, every element appears twice except for one. Find that single one.

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

使用位操作异或，异或能使相同值得位数为0，这样两两相同的数都被抵消了。

public class Solution {
    //Time: O(n)  Space: O(1)
    public int singleNumber(int[] A) {
        if (A == null || A.length == 0) {
            return -1;
        }
        
        int num = A[0];
        for (int i = 1; i < A.length; i++) {
            num = num ^ A[i];
        }
        return num;
    }
}  

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?

可以用一个O(32)的数组来记录每个digit出现次数，最后对3求余。
也可以像Single Number I，利用二进制模仿三进制，用两个变量one, two，当两者都是1的时候说明出现3次，置0。
所以我们可以用二进制来模拟各种进制

public class Solution {
    //Time: O(n)  Space: O(1)
    public int singleNumber(int[] A) {
        int one = 0;
        int two = 0;
        for (int i = 0; i < A.length; i++) {
            two = two | (one & A[i]);
            one ^= A[i];
            int tmp = ~(one & two);
            one = tmp & one;
            two = tmp & two;
        }
        return one;
    }
}