循环码排列

/*
 * The task is to generate a Gray code sequence of length 2^n where the first element is 'start'.
 * The solution here uses Java Streams to generate the sequence efficiently.
 */

import java.util.stream.Collectors;
import java.util.stream.IntStream;
import java.util.List;

public class Solution {
    public List<Integer> circularPermutation(int n, int start) {
        return IntStream.range(0, 1 << n)
                .map(i -> start ^ (i ^ (i >> 1)))
                .boxed()
                .collect(Collectors.toList());
    }
}