【算法】【线性表】【数组】Gray Code 格雷码

1  题目

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

2  解答

咦，首先这个题做之前得先看看什么是格雷码：https://baike.baidu.com/item/%E6%A0%BC%E9%9B%B7%E7%A0%81/6510858

在一组数的编码中，若任意两个相邻的代码只有一位二进制数不同，则称这种编码为格雷码（Gray Code）。

public class Solution {
    /**
     * @param n: a number
     * @return: Gray code
     */
    public List<Integer> grayCode(int n) {
        // write your code here
        // 定义返回结果
        List<Integer> res = new ArrayList<>();
        res.add(0);
        // 如果 n == 0,直接返回
        if (n == 0) {
            return res;
        }
        res.add(1);
        for (int i = 1; i < n; i++){
            // 逆序取，每个取出来然后跟1左移i位进行或运算
            int len = res.size();
            for (int j = len - 1; j >= 0; j--) {
                res.add(1 << i | res.get(j));
            }
        }
        return res;
    }
}

加油。