• 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[]

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  • 原文地址:https://www.cnblogs.com/EdwardLiu/p/3982823.html
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