# No bit manipulation required 3ms C++

• Go from the most left bit to the right.
• The number of grey code is 2^n
• Divide 2^n to half
• For the first half, the next bit's first half will be 0 and the second half will be 1.
• For the second half, the next bit's first half will be 1 and the second half will be 0.
``````class Solution {
private:
void rec(int start_i, int end_i, int zero_or_one_first, int power, vector<int>* result) {
if (power < 0) return;
int mid = (start_i + end_i) / 2;
int add = (int) pow(2, power);
if (zero_or_one_first == 0) {
for (int i = mid+1; i <= end_i; i++) (*result)[i] += add;
} else {
for (int i = start_i; i <= mid; i++) (*result)[i] += add;
}
rec(start_i, mid, 0, power-1, result);
rec(mid+1, end_i, 1, power-1, result);
}
public:
vector<int> grayCode(int n) {
vector<int> result((int) pow(2, n), 0);
rec(0, (int) pow(2, n) - 1, 0, n-1, &result);
return result;
}
};
``````

Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect.