LeetCode GrayCode

class Solution {
public:
    vector<int> grayCode(int n) {
        vector<int> res;
        res.push_back();

        long pow2 = ;
        for (int i=; i <= n; i++) {
            int old_len = res.size();
            for (int i=; i<old_len; i++) {
                res.push_back(pow2 + res[old_len - i - ]);
            }
            pow2<<=;
        }
        return res;
    }
};

再水一发， 不过n==0时，觉得应该返回一个空的vector

简化版：

class Solution {
public:
    vector<int> grayCode(int n) {
        vector<int> res;
        res.push_back();

        for (int i=; i<n; i++) {
            int idx = res.size() - ;
            while (idx >= ) {
                res.push_back(res[idx--] | <<i);
            }
        }

        return res;
    }
};

递归方法：

class Solution {
public:
    vector<int> grayCode(int n) {
        vector<int> res;
        if (n < ) {
            res.push_back();
            return res;
        } else if (n == ) {
            res.push_back();
            res.push_back();
            return res;
        }
        vector<int> sub = grayCode(n - );
        int len = sub.size();
        for (int i=; i < len; i++) {
            res.push_back(sub[i]);
        }
        for (int i=len-; i>=; i--) {
            res.push_back((<<(n-))|sub[i]);
        }
        return res;
    }
};