Leetcode: 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.

Analysis:

Try one more example, n = 3:

000 - 0

001 - 1

011 - 3

010 - 2

110 - 6

111 - 7

101 - 5

100 - 4 

Comparing n = 2: [0,1,3,2] and n=3: [0,1,3,2,6,7,5,4], we found that the first four numbers in case n=3 are the same as the the numbers in case n=4.  Besides, [6,7,5,4] = [2+4,3+4,1+4,0+4].  Which means remaining numbers in case n=3 can also be calculated from the numbers in case n=4 in reversing order.  Therefore, we decided to use recursive approach to form the resulting ArrayList.

 1 public class Solution {
 2     public ArrayList<Integer> grayCode(int n) {
 3         ArrayList<Integer> res = new ArrayList<Integer>();
 4         if(n == 0) {
 5             res.add(0);
 6         }
 7         else {
 8             ArrayList<Integer> pre = grayCode(n-1);
 9             res.addAll(pre);
10             for (int i=pre.size()-1; i>=0; i--) {
11                 res.add(pre.get(i) + (int)Math.pow(2, n-1));
12             }
13         }
14         return res;
15     }
16 }

Iterative 解法（推荐方法）：每次也不用定义一个 ArrayList<Integer> pre = res, 只用记录当前res 的size就好了

 1 public class Solution {
 2     public List<Integer> grayCode(int n) {
 3         List<Integer> res = new ArrayList<Integer>();
 4         if (n < 0) return res;
 5         res.add(0);
 6         if (n == 0) return res;
 7         for (int i=1; i<=n; i++) {
 8             int size = res.size();
 9             for (int j=size-1; j>=0; j--) {
10                 res.add(res.get(j) + (int)Math.pow(2, i-1));
11             }
12         }
13         return res;
14     }
15 }

 这些代码都要注意一个细节，即input为0时，output为[0]，并不是[]

转载于:https://www.cnblogs.com/EdwardLiu/p/3982823.html