leetcode 89. Gray Code 按照index递归DFS解决 + Grey码生成公式 + 位运算直接计算

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.

这个是学习信号与处理的时候讲到的格雷码，就是相邻的元素的数量差异最小。这个直接按照镜子的映射关系来设置就可以了。

代码如下：

import java.util.ArrayList;
import java.util.List;

/*
* 格雷码，就是按照index递归解决
* */
public class Solution
{
List<Integer> res=new ArrayList<>();
public List<Integer> grayCode(int n)
{
if(n<=0)
{
res.add(0);
return res;
}

List<List<Integer>> flag=new ArrayList<List<Integer>>();
ByReci(0,n,flag);
return res;
}
private void ByReci(int index, int n,List<List<Integer>> flag)
{
if(index==n)
{
for(int i=0;i<flag.size();i++)
{
List<Integer> one=flag.get(i);
int sum=0;
for(int j=0;j<one.size();j++)
sum+=one.get(j)*Math.pow(2, j);

res.add(sum);
}
}else
{
if(index==0)
{
List<Integer> one=new ArrayList<>();
one.add(0);
List<Integer> two=new ArrayList<>();
two.add(1);
flag.add(one);
flag.add(two);
ByReci(index+1,n, flag);
}else
{
for(int i=0;i<flag.size();i++)
flag.get(i).add(0);
for(int i=flag.size()-1;i>=0;i--)
{
List<Integer> one=new ArrayList<>(flag.get(i));
one.remove(one.size()-1);
one.add(1);
flag.add(one);
}
ByReci(index+1, n, flag);
}
}
}
}

下面是C++的做法，我是网上看到的做法，使用公式和位运算来计算完成的，很棒的做法，可以参考这个链接LeetCode — 89. Gray Code

代码如下：

#include <iostream>
#include <string>
#include <vector>

using namespace std;

/*
参考这个链接：http://blog.csdn.net/makuiyu/article/details/44926463
此题要求生成n位格雷码，共计有2^n个。
根据上面公式G(N) = B(n) XOR (B(n)/2)，即可逐步生成所有格雷码。
*/
class Solution
{
public:
vector<int> grayCode(int n)
{
vector<int> res;
int size = 1 << n;
for (int i = 0; i < size; i++)
res.push_back(i^(i>>1));
return res;
}
};