LeetCode每日一题(Gray Code)

最新推荐文章于 2023-01-23 18:05:40 发布

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本文介绍了一种生成n位灰码序列的有效方法，并提供了一个Rust语言实现的例子。灰码是一种特殊的二进制码序列，相邻两个码字仅有一位不同。文章详细解释了如何通过递归方法构建序列，同时附带了完整的代码示例。

An n-bit gray code sequence is a sequence of 2n integers where:

Every integer is in the inclusive range [0, 2n - 1],
The first integer is 0,
An integer appears no more than once in the sequence,
The binary representation of every pair of adjacent integers differs by exactly one bit, and
The binary representation of the first and last integers differs by exactly one bit.

Given an integer n, return any valid n-bit gray code sequence.

Example 1:

Input: n = 2
Output: [0,1,3,2]
Explanation:
The binary representation of [0,1,3,2] is [00,01,11,10].
- 00 and 01 differ by one bit
- 01 and 11 differ by one bit
- 11 and 10 differ by one bit
- 10 and 00 differ by one bit
[0,2,3,1] is also a valid gray code sequence, whose binary representation is [00,10,11,01].
- 00 and 10 differ by one bit
- 10 and 11 differ by one bit
- 11 and 01 differ by one bit
- 01 and 00 differ by one bit

Example 2:

Input: n = 1
Output: [0,1]

Constraints:

1 <= n <= 16

这题前段时间遇到过一回，费了半天劲也没做出来，最后放弃了，结果今天又蹦出来了，赤裸裸的挑衅，办它。

思路:

这题在我看来就是纯粹的找规律的题，拿出纸笔，写出一组答案，然后开始相面。

n = 3时答案为[0, 1, 3, 2, 6, 7, 5, 4], 转换成二进制就是[000, 001, 011, 010, 110, 111, 101, 100]。

端详了半天，一开始以为是用两层循环就好了，后来发现根本没法覆盖所有的数，于是继续找规律，最终在mask身上找到了突破口。我们每生成一个新的数字，就是在原有数字的基础上^(1 << k)得来，这个1<<k就是mask，我们把所有用到的mask列出来：

变化 Mask left shift

000 -> 001: 001 0

001 -> 011: 010 1

011 -> 010: 001 0

010 -> 110: 100 2

110 -> 111: 001 0

111 -> 101: 010 1

101 -> 100: 001 0

貌似也没什么规律是不是，一开始我认为是(0), (1, 0), (2, 0, 1)，(3, 0, 1, 2)， ......, (n, 0, 1, 2, ..., n-1), 这种规律，后来实现的时候发现不对，因为n=3的时候会缺少最后一个元素。后来又一想，有没有可能是遍历嵌套递归呢？试了一下，通过了。

代码(Rust):

impl Solution {
    fn rc(n: i32, num: &mut i32, l: &mut Vec<i32>) {
        *num = *num ^ (1 << n);
        l.push(*num);
        for i in 0..n {
            Solution::rc(i, num, l);
        }
    }
    pub fn gray_code(n: i32) -> Vec<i32> {
        let mut ans: Vec<i32> = vec![0];
        let mut num = 0;
        for i in 0..n {
            Solution::rc(i, &mut num, &mut ans)
        }
        ans
    }
}