Gray Code 格雷码 @LeetCode

没有思路，后来看了http://www.cnblogs.com/lihaozy/archive/2012/12/31/2840437.html才知道原来可以把binary code直接转为gray code：

Binary Code ：1011 要转换成Gray Code

　　1011 = 1（照写第一位）, 1(第一位与第二位异或 1^0 = 1), 1(第二位异或第三位， 0^1=1), 0 (1^1 =0) = 1110

　　其实就等于 (1011 >> 1) ^ 1011 = 1110

有了上面的等式写code就简单了

package Level4;

import java.util.ArrayList;

/**
 *  Gray Code
 *  
 *  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.
 *
 */
public class S89 {

	public static void main(String[] args) {

	}

	public static ArrayList<Integer> grayCode(int n) {
		int size = 1 << n;		// 最大size，相当于2^n
		ArrayList<Integer> ret = new ArrayList<Integer>();
		
		for(int i=0; i<size; i++){
			// 右移一位后异或
			ret.add(i ^ (i>>1));
		}
		return ret;
	}

}

其实这道题的正解是找规律：

以3位格雷码为例。
0 0 0
0 0 1
0 1 1
0 1 0
1 1 0
1 1 1
1 0 1
1 0 0
可以看到第n位的格雷码由两部分构成，一部分是n-1位格雷码，再加上 1<<(n-1)和n-1位格雷码的逆序的和。

public class Solution {  
    public ArrayList<Integer> grayCode(int n) {  
        if(n==0) {  
            ArrayList<Integer> result = new ArrayList<Integer>();  
            result.add(0);  
            return result;  
        }  
          
        ArrayList<Integer> tmp = grayCode(n-1);  
        int addNumber = 1 << (n-1);  
        ArrayList<Integer> result = new ArrayList<Integer>(tmp);  
        for(int i=tmp.size()-1;i>=0;i--) {  
            result.add(addNumber + tmp.get(i));  
        }  
        return result;  
    }  
}  

http://blog.youkuaiyun.com/worldwindjp/article/details/21536103