Medium
Given 2 integers n
and start
. Your task is return any permutation p
of (0,1,2.....,2^n -1)
such that :
p[0] = start
p[i]
and p[i+1]
differ by only one bit in their binary representation.p[0]
and p[2^n -1]
must also differ by only one bit in their binary representation.Example 1:
Input: n = 2, start = 3
Output: [3,2,0,1]
Explanation: The binary representation of the permutation is (11,10,00,01). All the adjacent element differ by one bit. Another valid permutation is [3,1,0,2]
Example 2:
Input: n = 3, start = 2
Output: [2,6,7,5,4,0,1,3]
Explanation: The binary representation of the permutation is (010,110,111,101,100,000,001,011).
Constraints:
1 <= n <= 16
0 <= start < 2 ^ n
import java.util.ArrayList;
import java.util.List;
public class Solution {
public List<Integer> circularPermutation(int n, int start) {
List<Integer> l1 = new ArrayList<>();
for (int i = 0; i < (1 << n); i++) {
l1.add(start ^ (i ^ (i >> 1)));
}
return l1;
}
}