binary num　　　　         gray code
　0=000　　　　　　000 = 000 ^ (000>>1)
　1=001　　　　　　001 = 001 ^ (001>>1)
　2=010　　　　　　011 = 010 ^ (010>>1)
　3=011　　　　　　010 = 011 ^ (011>>1)
　...
　7=111　　　　　　100 = 111 ^ (111>>1)

class Solution {
public:
    vector<int> grayCode(int n) {
        if(n<0) return vector<int>(0);
        
        vector<int> ans;
        int sum = pow(2, n);
        for(unsigned int i=0; i<sum; i++)
        {
            ans.emplace_back(i^(i>>1));
        }
        
        return ans;
    }
};
 

2、My idea is to generate the sequence iteratively. For example, when n=3, we can get the result based on n=2.
00,01,11,10 -> (000,001,011,010 ) (110,111,101,100). The middle two numbers only differ at their highest bit, while the rest numbers of part two are exactly symmetric of part one. It is easy to see its correctness.
Code is simple:

public List<Integer> grayCode(int n) {
    List<Integer> rs=new ArrayList<Integer>();
    rs.add(0);
    for(int i=0;i<n;i++){
        int size=rs.size();
        for(int k=size-1;k>=0;k--)
            rs.add(rs.get(k) | 1<<i);
    }
    return rs;
}