Leetcode(89) Gray Code

发表于 | 分类于

Description

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.

Example 1:

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Input: 2
Output: [0,1,3,2]
Explanation:
00 - 0
01 - 1
11 - 3
10 - 2

For a given n, a gray code sequence may not be uniquely defined.
For example, [0,2,3,1] is also a valid gray code sequence.

00 - 0
10 - 2
11 - 3
01 - 1

Example 2:

1
2
3
4
5
Input: 0
Output: [0]
Explanation: We define the gray code sequence to begin with 0.
             A gray code sequence of n has size = 2n, which for n = 0 the size is 20 = 1.
             Therefore, for n = 0 the gray code sequence is [0].

解法

题意是每次只允许变动一位，根据给出的位数得到序列。我采用的是最原始的深度优先搜索算法，利用hash表不能重复的性质解这道题，每次在写深度优先搜索的时候注意回溯的处理。

具体代码如下：

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
class Solution {
    List<Integer> res = new ArrayList<>();
    HashSet<Integer> hs = new HashSet<>();

    public List<Integer> grayCode(int n) {
        int[] s = new int[n];
        for (int i = 0; i < n; i++) {
            s[i] = 0;
        }
        res.add(0);
        hs.add(0);
        helper(s);
        return res;
    }

    void helper(int[] s) {
        if (hs.size() >= Math.pow(2, s.length)) {
            return;
        }
        int n = s.length;
        for (int i = 0; i < n; i++) {
            s[i] = 1 - s[i];
            int t = trans(s);
            if (!hs.contains(t)) {
                hs.add(t);
                res.add(t);
                if (hs.size() >= Math.pow(2, s.length)) {
                    return;
                }
                helper(s);
            }
            s[i] = 1 - s[i];
        }
    }

    int trans(int[] s) {
        int res = 0;
        for (int i = 0; i < s.length; i++) {
            res += (s[i] * Math.pow(2, i));
        }
        return res;
    }
}