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.
思路:
初次看到的时候觉得可能是要做图搜索的样子。但是事实上并不那么复杂,只要找到下一个未访问过的数就可以了。这里用一个简单的异或操作就可以flip一个位:x ^= (1 << bit)。另外用了一个set来保存已经访问过的数字。set一般是通过hash的方法来搜索的,因此平均花费的时间是O(1)。而map的key搜索则要O(log n),因为map的实现通常是基于红黑树的。
题解:
class Solution { public: vector<int> grayCode(int n) { vector<int> ret; set<int> usedval; auto flip = [&](int k, int bit) -> int { return k ^ (1 << bit); }; queue<int> visit; visit.push(0); while(!visit.empty()) { auto front = visit.front(); visit.pop(); ret.push_back(front); usedval.insert(front); // now check possible choices // we have to avoid the sgn bit for(int i = 0; i < n; ++i) { int newval = flip(front, i); if (usedval.find(newval) != usedval.end()) continue; else { visit.push(newval); break; } } } return ret; } };