leetcode 89. Gray Code 按照index递归解决 + Grey码生成公式

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;    }};