【Leetcode】Gray Code

最新推荐文章于 2025-09-09 09:10:31 发布

yyz20002008

最新推荐文章于 2025-09-09 09:10:31 发布

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文章标签： java leetcode List arraylist

本文介绍了一种生成格雷码序列的有效算法，并提供了多种实现方式，包括递归和位操作方法，适用于理解与实现格雷码的核心逻辑。

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:

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.

这道题需要很熟练位运算

Java:

1.http://www.cnblogs.com/feiling/p/3252084.html

二进制码->格雷码（编码）：从最右边一位起，依次将每一位与左边一位异或（XOR），作为对应格雷码该位的值，最左边一位不变（相当于左边是0）；
格雷码->二进制码（解码）：从左边第二位起，将每位与左边一位解码后的值异或，作为该位解码后的值（最左边一位依然不变）。

public class Solution {
    public List<Integer> grayCode(int n) {
         
        int size = 1<<n;//2 to n power
        ArrayList<Integer> res = new ArrayList<Integer>();
        for(int i =0;i<size;i++)
            res.add(i ^(i>>1));//this is how they get gray number, don't compare result with i; 
        return res;
    
    }
}

2. 也是一种形式的倒序+2的n-1次构成方法

 public class Solution {
    public List<Integer> grayCode(int n) {
         // Start typing your Java solution below
        // DO NOT write main() function
       ArrayList<Integer> ans=new ArrayList<Integer>();
       ans.add(0);
       if(n==0) {return ans;}
       for(int i=0;i<n;i++){
           int m=(int)Math.pow(2, i);
           for(int j=m-1;j>=0;j--){
               ans.add(ans.get(j)+m);在每一套格雷码的末尾开始加2的i次幂
           }
       }
       return ans;
    
    }
}

3.http://blog.youkuaiyun.com/u010500263/article/details/18209669

public class Solution {  
    public ArrayList<Integer> grayCode(int n) {  
        ArrayList<Integer> res = new ArrayList<Integer>();  
        if(n==0) {  
            res.add(0);  
            return res;  
        }  
          
        ArrayList<Integer> preRes = grayCode(n-1);  
        res.addAll(preRes);  
        for(int i=preRes.size()-1; i>=0; i--){  
            res.add(preRes.get(i)+(int)Math.pow(2, n-1));  
        }  
        return res;  
    }  
}

4. http://blog.youkuaiyun.com/linhuanmars/article/details/24511221

public ArrayList<Integer> grayCode(int n) {
    ArrayList<Integer> res = new ArrayList<Integer>();
    if(n<0)
        return res;
    if(n==0)
    {
        res.add(0);
        return res;
    }
    res.add(0);
    res.add(1);
    for(int i=2;i<=n;i++)
    {
        int size = res.size();
        for(int j=size-1;j>=0;j--)
        {
            res.add(res.get(j)+(1<<(i-1)));
        }
    }
    return res;
}