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.

解答：

1.我的想法。

先是加入0，1。 然后对0,1从前往后的后面加0，再对0，1进行从后往前的加1.即 00，10，11，01.

以此类推，再在以上四个里面，先是从前往后挨个在最后加0，再是从后往前挨个加1.

class Solution {
public:
vector<int> grayCode(int n) {
vector<vector<int>> res,vec;
vector<int> v;
v.push_back(0);
if(n==0)
return v;
vec.push_back(v);
v.push_back(1);
if(n == 1)
return v;
v.clear();
v.push_back(1);
vec.push_back(v);
for(int i = 1; i < n; i++){
res.clear();
for(auto& j:vec){
j.push_back(0);
res.push_back(j);
j.pop_back();
}
for(int i = vec.size()-1; i >=0; i--){
vec[i].push_back(1);
res.push_back(vec[i]);
}
vec.clear();
vec.insert(vec.end(),res.begin(),res.end());
}
v.clear();

for(int k = 0; k < res.size(); k++){
int sum = 0;
for(int p = res[k].size()-1; p >=0 ; p--){
sum = sum * 2 +res[k][p];
}
v.push_back(sum);
}
return v;
}
};

2.别人的代码

方法跟我一样，只不过他很巧妙的利用了移位。

vector<int> grayCode(int n) {
vector<int> res;
int bits = 1;
int ntmp=0;
res.push_back(0);
for(int i=0;i<n;i++)
{
bits=1<<i;
for(int j = res.size()-1;j>=0;j--){
ntmp = res.at(j)+bits;
res.push_back(ntmp);
}

}
return res;
}