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.
divide && conquer
public List<Integer> grayCode(int n) { List<Integer> res = new ArrayList<Integer>(); if(n == 0) { res.add(0); } else { List<Integer> pre = grayCode(n-1); res.addAll(pre); for (int i=pre.size()-1; i>=0; i--) { res.add(pre.get(i) + (int)Math.pow(2, n-1)); } } return res; }
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
public class Solution { public List<Integer> grayCode(int n) { List<Integer> res = new ArrayList<Integer>(); if (n < 0) return res; res.add(0); if (n == 0) return res; for (int i=1; i<=n; i++) { int size = res.size(); for (int j=size-1; j>=0; j--) { res.add(res.get(j) + (int)Math.pow(2, i-1)); } } return res; } }
算法复杂度是O(2+2^2+...+2^n-1)=O(2^n),所以是指数量级的,因为是结果数量无法避免。空间复杂度则是结果的大小,也是O(2^n)。
这些代码都要注意一个细节,即input为0时,output为[0],并不是[]