Gray Code 题解

题目来源：Gray Code

> 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, print the sequence of gray code. A gray code sequence must begin with 0. For example, given n = 2, return [0,1,3,2]. Its gray code sequence is: 00 - 0 01 - 1 11 - 3 10 - 2 Note: For a given n, a gray code sequence is not uniquely defined. For example, [0,2,3,1] is also a valid gray code sequence according to the above definition. For now, the judge is able to judge based on one instance of gray code sequence. Sorry about that.

解题思路：

逆序

注意观察，n每增加1，即是在n-1的结果之上，最高位加1，并按照n-1的逆序。

n = 1 
0 
1 
n=2 
0 0 
0 1 
--- 
1 1 
1 0 
n=3 
0 0 0 
0 0 1 
0 1 1 
0 1 0 
------ 
1 1 0 
1 1 1 
1 0 1 
1 0 0 

    vector<int> grayCode(int n) 
    {
        assert(n>=0);
        if(n == 0) return std::move(vector<int>(1,0));
        vector<int> result;
        result.push_back(0); result.push_back(1);
        for(int i = 1; i < n; i++)
        {
            int len = result.size();
            for(int j = len-1; j >=0; j--)
                result.push_back(result[j] + (1<<i));
        }
        return move(result);
    }

公式法

格雷码

G：格雷码  B：二进制码 
G(N) = (B(n)/2) XOR B(n) 
Binary Code(1011)要转换成Gray Code = (1011 >> 1) ^ 1011 = 1110 

    vector<int> grayCode(int n) 
    {
        assert(n>=0);
        int len = 1 << n;//std::pow(2,n);
        vector<int> result(len, 0);
        for(int i = 1; i < len; i++)
            result[i] = (i>>1)^i;
        return move(result);
    }