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[]