題目

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]

思路

• n = 0 [0]
• n = 1 [0, 1]
• n = 2 [0, 1, 3, 2]
• n = 3 [0, 1, 3, 2, 6, 7, 5, 4]

認真找尋規律，每一個都會是前一個的 reverse 再加上最大的 bit，也就是 2^(n - 1)

Solution

/**
 * @param {number} n
 * @return {number[]}
 */
var grayCode = function(n) {
    var res = [0]
    var count = 0
    while(count < n) {
        const offset = Math.pow(2, n - 1)
        const temp = [...res].reverse().map(v => v + offset)
        res = res.concat(temp)
        n--
    }
    return res
};

Complexity

• Time complexity : O(n)
• Space complexity : O(1)