In this Leetcode Gray Code problem solution An n-bit Gray code sequence is a sequence of 2n integers where:

1. Every integer is in the inclusive range [0, 2n - 1],
2. The first integer is 0,
3. An integer appears no more than once in the sequence,
4. The binary representation of every pair of adjacent integers differs by exactly one bit, and
5. The binary representation of the first and last integers differs by exactly one bit.

Given an integer n, return any valid n-bit gray code sequence.

## Problem solution in Python.

```class Solution:
def grayCode(self, n: int) -> List[int]:
ans = []
for i in range(0, 2 ** n):
b = bin(i)[2:]
temp = b
for j in range(1, len(b)):
temp += str(int(b[j - 1]) ^ int(b[j]))
ans.append(temp)

return list(map(lambda x: int(x, 2), ans))
```

## Problem solution in Java.

```class Solution {
public List<Integer> grayCode(int n) {
ArrayList<Integer> res = new ArrayList<>();
int num = 1<<n;
for(int i = 0; i < num; i++) {
}
return res;
}
}
```

## Problem solution in C++.

```vector<int> grayCode(int n) {
if(n == 0) return {0};
vector<int> ret = {0, 1};

for(int i=2; i<=n; i++)
for(int j=ret.size()-1; j>=0; j--)
ret.push_back(ret[j]+(1<<i-1));

return ret;
}
```

## Problem solution in C.

```int* grayCode(int n, int* returnSize) {
*returnSize=1<<n;
unsigned int* pAns=(unsigned int*)malloc(*returnSize*sizeof(unsigned int));
pAns=0;
for (unsigned int i=1;i<*returnSize;i++){
unsigned int temp=i;
for (int j=0;j<n;j++){//xor the last bit which is not zero
if (temp%2!=0){
pAns[i]=pAns[i-1]^(1<<j);
break;
}
temp=temp>>1;
}
}
return pAns;
}
```