[LintCode] Gray Code 格雷码

 

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, find the sequence of gray code. A gray code sequence must begin with 0 and with cover all 2nintegers.

 Notice

For a given n, a gray code sequence is not uniquely defined.

[0,2,3,1] is also a valid gray code sequence according to the above definition.

Have you met this question in a real interview? 

Yes

Example

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

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

Challenge 

O(2ⁿ) time.

 

LeetCode上的原题，请参见我之前的博客Gray Code。

 

解法一：

class Solution {
public:
    /**
     * @param n a number
     * @return Gray code
     */
    vector<int> grayCode(int n) {
        vector<int> res;
        for (int i = 0; i < pow(2, n); ++i) {
            res.push_back((i >> 1) ^ i);
        }
        return res;
    }
};

 

解法二：

class Solution {
public:
    /**
     * @param n a number
     * @return Gray code
     */
    vector<int> grayCode(int n) {
        vector<int> res{0};
        for (int i = 0; i < n; ++i) {
            int size = res.size();
            for (int j = size - 1; j >= 0; --j) {
                res.push_back(res[j] | (1 << i));
            }
        }
        return res;
    }
};

 

解法三：

class Solution {
public:
    /**
     * @param n a number
     * @return Gray code
     */
    vector<int> grayCode(int n) {
        vector<int> res{0};
        int len = pow(2, n);
        for (int i = 1; i < len; ++i) {
            int pre = res.back();
            if (i % 2 == 1) {
                pre = (pre & (len - 2)) | (~pre & 1);
            } else {
                int cnt = 1, t = pre;
                while ((t & 1) != 1) {
                    ++cnt; t >>= 1;
                }
                if ((pre & (1 << cnt)) == 0) pre |= (1 << cnt);
                else pre &= ~(1 << cnt);
            }
            res.push_back(pre);
        }
        return res;
    }
};

 

解法四：

class Solution {
public:
    /**
     * @param n a number
     * @return Gray code
     */
    vector<int> grayCode(int n) {
        vector<int> res{0};
        unordered_set<int> s;
        stack<int> st;
        s.insert(0);
        st.push(0);
        while (!st.empty()) {
            int t = st.top(); st.pop();
            for (int i = 0; i < n; ++i) {
                int k = t;
                if ((k & (1 << i)) == 0) k |= (1 << i);
                else k &= ~(1 << i);
                if (s.count(k)) continue;
                s.insert(k);
                st.push(k);
                res.push_back(k);
                break;
            }
        }
        return res;
    }
};

 

解法五：

class Solution {
public:
    /**
     * @param n a number
     * @return Gray code
     */
    vector<int> grayCode(int n) {
        vector<int> res;
        unordered_set<int> s;
        helper(n, s, 0, res);
        return res;
    }
    void helper(int n, set<int>& s, int out, vector<int>& res) {
        if (!s.count(out)) {
            s.insert(out);
            res.push_back(out);
        }
        for (int i = 0; i < n; ++i) {
            int t = out;
            if ((t & (1 << i)) == 0) t |= (1 << i);
            else t &= ~(1 << i);
            if (s.count(t)) continue;
            helper(n, s, t, res);
            break;
        }
    }
};