[LeetCode]Gray Code

class Solution {
//write down some cases, and try to find out the regular pattern
//gray code is not unique, so we should find out at least one regular pattern
//one of these regular pattern is that:
//let f(i) be the gray code sequence when n=i,
//then the number of sequence f(i+1) is twice of f(i), we consider f(i) is a half sequence of f(i+1)
//because the sequence f(i) is valid, so another half valid sequence can be find to add 1 in the highest bit of each element of f(i), 
//the last thing we need to figure out is that we should concatenate these two sequence, the last one element of the second half will be 
//only one digit different from the last of the first half, so we just need to reverse the second half
public:
	vector<int> grayCode(int n) {
		// Start typing your C/C++ solution below
		// DO NOT write int main() function
		vector<int> ans;
		ans.push_back(0);
		for (int i = 1; i <= n; ++i)
		{
			int size = ans.size();
			for (int j = size-1; j >= 0; --j)
				ans.push_back( ans[j]^(1<<(i-1)) );
		}
		return ans;
	}
};

second time

class Solution {
public:
    vector<int> grayCode(int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<int> ans(1);
        ans[0] = 0;
        int add = 1;
        for(int i = 1; i <= n; ++i)
        {
            int oldSize = ans.size();
            for(int j = oldSize-1; j >= 0; --j) ans.push_back(ans[j]+add);
            add *= 2;
        }
        return ans;
    }
};