[LintCode] Gray Code

Problem

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, find the sequence of gray code. A gray code sequence must begin with 0 and with cover all 2n integers.
http://mathworld.wolfram.com/...

Example

Given n = 2, return [0,1,3,2]. Its gray code sequence is:

00 - 0
01 - 1
11 - 3
10 - 2

Solution

其实就是找规律，0-01-0132－01326754这样，每个循环(i)：

1. 增加当前res.size()个新数；

2. 新的循环先在2进制的第(i+1)位加1，同时与之前res所存的数字倒序相加产生新数；

3. 存入新数，进入下一个循环后更新size。

参考了codesolutiony的算法，简化了一下：

public class Solution {
    public ArrayList<Integer> grayCode(int n) {
        ArrayList<Integer> res = new ArrayList<Integer>();
        res.add(0);
        for (int i = 0; i < n; i++) {
            //每个循环更新size: 1, 3, 7...
            int size = res.size() - 1;
            //i对应最高位是第i+1位，res就增加2^(i+1)个数：(1<<i)+res.get(j)
            //j为倒序循环(size+1)次，会镜像提取上一次循环产生的结果
            //01 - 32: 镜像＋2  0132 - 6754: 镜像＋4 
            //0 1 3 2 6 7 5 4 - 12 13 15 14 10 11 9 8: 镜像+8
            for (int j = size; j >= 0; j--) {
                res.add((1<<i) + res.get(j));
            }
        }
        return res;
    }
}

Recursion

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

Using XOR

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