# 89. 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.

``````Int    Grey Code    Binary
0  　　  000        000
1  　　  001        001
2   　 　011        010
3   　 　010        011
4   　 　110        100
5   　 　111        101
6   　 　101        110
7   　　 100        111
``````

``````// Binary to grey code
class Solution {
public:
vector<int> grayCode(int n) {
vector<int> res;
for (int i = 0; i < pow(2,n); ++i) {
res.push_back((i >> 1) ^ i);
}
return res;
}
};
``````

``````// Mirror arrangement
class Solution {
public:
vector<int> grayCode(int n) {
vector<int> res{0};
for (int i = 0; i < n; ++i) {
int size = res.size();
for (int j = size - 1; j >= 0; --j) {
res.push_back(res[j] | (1 << i));
}
}
return res;
}
};
``````

0 0 0
0 0 1
0 1 1
0 1 0
1 1 0
1 1 1
1 0 1
1 0 0

``````// Direct arrangement
class Solution {
public:
vector<int> grayCode(int n) {
vector<int> res{0};
int len = pow(2, n);
for (int i = 1; i < len; ++i) {
int pre = res.back();
if (i % 2 == 1) {
pre = (pre & (len - 2)) | ((~pre) & 1);
} else {
int cnt = 1, t = pre;
while ((t & 1) != 1) {
++cnt;
t >>= 1;
}
if ((pre & (1 << cnt)) == 0) pre |= (1 << cnt);
else pre &= ~(1 << cnt);
}
res.push_back(pre);
}
return res;
}
};
``````

``````class Solution {
public:
vector<int> grayCode(int n) {
vector<int> res;
unordered_set<int> s;
helper(n, s, 0, res);
return res;
}
void helper(int n, unordered_set<int>& s, int out, vector<int>& res) {
if (!s.count(out)) {
s.insert(out);
res.push_back(out);
}
for (int i = 0; i < n; ++i) {
int t = out;
if ((t & (1 << i)) == 0) t |= (1 << i);
else t &= ~(1 << i);
if (s.count(t)) continue;
helper(n, s, t, res);
break;
}
}
};
``````

``````class Solution {
public:
vector<int> grayCode(int n) {
vector<int> res{0};
unordered_set<int> s;
stack<int> st;
st.push(0);
s.insert(0);
while (!st.empty()) {
int t = st.top(); st.pop();
for (int i = 0; i < n; ++i) {
int k = t;
if ((k & (1 << i)) == 0) k |= (1 << i);
else k &= ~(1 << i);
if (s.count(k)) continue;
s.insert(k);
st.push(k);
res.push_back(k);
break;
}
}
return res;
}
};
``````

https://discuss.leetcode.com/topic/8557/an-accepted-three-line-solution-in-java

LeetCode All in One 题目讲解汇总(持续更新中…)

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