2.1.19 Gray Code

Link: https://oj.leetcode.com/problems/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 - 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.

我的思路：only remember output list 的前一半和后一半是对称的。生成了前一半，后一半也可以相应生成。前一半可以由grayCode(n-1)生成。

我的代码：基本一次过。可以不再做。

Note: 

1 Math.pow(2, n-1) 结果是double, 要强制类型转换成（int). 用1<<(n-1) 表示2^(n-1)更好。

2 basecase 是n=0, 不是n=1, 否则会出现Runtime error.

Approach I: Reflex-and-Prefix method

Time: O(2^n), Space: O(1)

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

A better code:

public class Solution {    public ArrayList<Integer> grayCode(int n) {        //solution2: Reflect and prefix        if(n == 0){            ArrayList<Integer> result = new ArrayList<Integer>();            result.add(0);//why? @4:57pm @6.22.2014            return result;        }        ArrayList<Integer> tmp = grayCode(n-1);        int add = 1 << (n-1);         ArrayList<Integer> result = new ArrayList<Integer>(tmp);        for(int i = tmp.size()-1; i>=0; i--){            result.add(add + tmp.get(i));        }        return result;    }}

Approach II: Math//don't understand

Time: O(2^n), Space: O(1)

public class Solution {//my 2nd after reading http://blog.csdn.net/xudli/article/details/8661913//However, I don't understand why this works    public ArrayList<Integer> grayCode(int n) {        ArrayList<Integer> result = new  ArrayList<Integer>();        int sz = 1<<n;        for(int i = 0; i < sz; i++){            result.add(i^(i>>1));        }        return result;    }}