LeetCode GrayCode

class Solution {
public:
    vector<int> grayCode(int n) {
        vector<int> res;
        res.push_back(0);

        long pow2 = 1;
        for (int i=1; i <= n; i++) {
            int old_len = res.size();
            for (int i=0; i<old_len; i++) {
                res.push_back(pow2 + res[old_len - i - 1]);
            }
            pow2<<=1;
        }
        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(0);

        for (int i=0; i<n; i++) {
            int idx = res.size() - 1;
            while (idx >= 0) {
                res.push_back(res[idx--] | 1<<i);
            }
        }
        
        return res;
    }
};

递归方法：

class Solution {
public:
    vector<int> grayCode(int n) {
        vector<int> res;
        if (n < 1) {
            res.push_back(0);
            return res;
        } else if (n == 1) {
            res.push_back(0);
            res.push_back(1);
            return res;
        }
        vector<int> sub = grayCode(n - 1);
        int len = sub.size();
        for (int i=0; i < len; i++) {
            res.push_back(sub[i]);
        }
        for (int i=len-1; i>=0; i--) {
            res.push_back((1<<(n-1))|sub[i]);
        }
        return res;
    }
};