思路:例举grey code序列,并找规律 :
n = 0: 0 n = 1: 0, 1 n = 2: 00, 01, 11, 10 (0, 1, 3, 2) n = 3: 000, 001, 011, 010, 110, 111, 101, 100 (0, 1, 3, 2, 6, 7, 5, 4) 以n = 3为例,grey code中前4个包括了n = 2的所有gray code。后4个则是前4个逆序后加上2^2。 推广:n = i的grey code的前一半包括了n = i-1的所有grey code,而后一半则为前一半逆序后家上2^(i-1)。 public class Solution { public List<Integer> grayCode(int n) { List<Integer> result = new LinkedList<>(); if(n<0) return result; result.add(0); int inc = 1; for (int i = 0; i < n; i++) { int size = result.size(); for(int j=size-1;j>=0;j--) { result.add(result.get(j)+inc); } inc = inc<<1; } return result; } }
解法二:O(n)
public class Solution { public List<Integer> grayCode(int n) { List<Integer> result = new LinkedList<>(); for (int i = 0; i < 1<<n; i++) result.add(i ^ i>>1); //1<<n = 2^n return result; } }
reference:
http://bangbingsyb.blogspot.com/2014/11/leetcode-gray-code.html
转载于:https://www.cnblogs.com/hygeia/p/5180874.html