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.

 

 解法1 dp＋非位运算

  规律：

public class Solution {    public List
     
       grayCode(int n) {        if(n == 0) {            List
      
        l1 = new ArrayList<>();            l1.add(0);            return l1;        }        List
       
        
         > list = new ArrayList<>();        List
         
           l1 = new ArrayList<>();        List
          
            res = new ArrayList<>(); l1.add("0"); l1.add("1"); list.add(l1); for(int i = 1 ; i < n ; i++){ List
           
             member = list.get(i-1); List
            
              next = new ArrayList<>(); for(String s : member){ String newS = "0"+ s; next.add(newS); } for(int j = member.size() -1 ; j >= 0; j--){ String newS = "1"+ member.get(j); next.add(newS); } list.add(next); } List
             
               l = list.get(n-1); for(String str : l){ res.add(getValueofBit(str)); } return res; } public int getValueofBit(String s){ int res = 0; for(int i = 0 ; i < s.length() ; i++){ res += Math.pow(2, s.length()-1-i) * (s.charAt(i) -'0'); } return res; }}
             
            
           
          
         
        
       
      
     

bit运算

 

public class Solution {    public List
     
       grayCode(int n) {        int count = (int) Math.pow(2, n);        List
      
        res = new ArrayList<>();        for(int i = 0; i < count ; i++){            res.add(i ^ (i >> 1));        }        return res;    }}