Gray Code

题目：

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.

就是把n位二进制数都转成十进制返回。代码：

 1     vector<int> grayCode(int n) {
 2         vector<int> result;
 3         if(!n) {
 4             result.push_back(0);
 5             return result;
 6         }
 7         bool* bits = (bool*)malloc(n);
 8         memset(bits,0,n);
 9         getGrayCode(result,n,0,bits);
10         return result;
11     }
12     void getGrayCode(vector<int>& result,int n,int i,bool* bits){
13         if(n==i){
14             result.push_back(b2i(bits,n));
15             return ;
16         }
17         getGrayCode(result,n,i+1,bits);
18         bits[i]=!bits[i];
19         getGrayCode(result,n,i+1,bits);
20     }
21     int b2i(bool* bits,int n){
22         int result=0,c=0;
23         for(int i=n-1;i>=0;i--){
24             if(!c) c=1;
25             else c*=2;
26             if(bits[i]){
27                 result += c;
28             }
29         }
30         return result;
31     }

时间复杂度太大了，以为过不了，结果还是过了，都2ⁿ*n了。不过规模已经是指数级，没办法降啊。
后来看到这个解法，于是去wikipedia上面看了下，确实非常简单，下面的链接：
http://blog.csdn.net/doc_sgl/article/details/12251523