[LeetCode] Gray Code

Link : Gray Code

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.

Analysis:
Write by hand 00, 01, 11, 10, 110, 111, 101, 100
The position changes is at  0,1,0,2,0,1,0
You find something?
Yes, it's symmetric!
The only thing we need to do is get the bit position as vector, and change bit according to it!
The bit_to_change is generated  by   A(n) = A(n-1) + n + A(n-1), by concatenation.

Yeah, later I found that there is a formula to generate Gray Code for each number as Gray(n) = (n>>1)^n.
But I still find my method is more interesting, LOL.

Code says!

 class Solution {  
 public:  
   std::vector<int> grayCode(int n) {  
     std::vector<int> ret;  
     std::vector<int>& bit_to_chage = get_bit_to_change(n-1);  
     int num = 0;  
     ret.push_back(num);  
     for(int i=0; i<bit_to_chage.size(); i++)  
     {  
      num ^= 1<<bit_to_chage[i];  
      ret.push_back(num);  
     }  
     return ret;  
   }  
   std::vector<int> get_bit_to_change(int n)  
   {  
    std::vector<int> ret;  
    if(n<0)   
      return ret;  
    std::vector<int>& v = get_bit_to_change(n-1);  
    ret.insert(ret.end(), v.begin(), v.end());  
    ret.push_back(n);  
    ret.insert(ret.end(), v.begin(), v.end());  
    return ret;  
   }  
 };