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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0001_0100.s0089_gray_code;

// #Medium #Math #Bit_Manipulation #Backtracking
// #2022_06_20_Time_3_ms_(98.59%)_Space_47.3_MB_(99.65%)

import java.util.Arrays;
import java.util.List;

/**
 * 89 - Gray Code\.
 *
 * Medium
 *
 * 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`
**/
public class Solution {
    public List grayCode(int n) {
        Integer[] n1 = {0};
        int shift = 1;
        while (n > 0) {
            Integer[] temp = new Integer[n1.length * 2];
            int pos = 0;
            for (Integer integer : n1) {
                temp[pos++] = integer;
            }
            for (int i = n1.length - 1; i >= 0; i--) {
                temp[pos++] = n1[i] | shift;
            }
            n1 = temp;
            shift <<= 1;
            n--;
        }
        return Arrays.asList(n1);
    }
}