LeetCode:Gray Code

Gray Code

Total Accepted: 50553 Total Submissions: 145538 Difficulty: Medium

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 - 001 - 111 - 310 - 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.

Subscribe to see which companies asked this question

Hide Tags

 Backtracking

格雷码：

上例中n=2，规律还不是很明显。来看看n=4时：

可以注意到，两个画框部分是对称的，只是前缀不同。当然上框部分再细分依然会找到这种规律。

code:

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

15年秋腾讯校招机试出现此题，题目要求用递归实现。

代码：

class Solution {public:vector<int> grayCode(int n) {vector<int> gray;gray.push_back(0);grayCode(gray, 0, n);return gray;}void grayCode(vector<int> &v,int s, int n){int size = v.size();if (s == n) return;for (int i = size - 1; i >= 0; i--){int code = (1 << s) | v[i];ivec.push_back(code);}grayCode(v, s+1, n);}};

根据百度百科中的介绍：

gray[i] = i ^ (i>>1)

因此有：

class Solution {public:    vector<int> grayCode(int n) {        vector<int> gray;    gray.push_back(0);    for(int i=0;i<size;i++) gray[i] = i ^ (i>>1);    return gray;    }};