leetcode-1238-Circular-Permutation-in-Binary-Representation

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:

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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:

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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

题目所求序列是一个 格雷码 序列, 相邻的二进制只有一位不同(有效位数), 首尾同样只有一位不同题目虽然给出了开始位置, 但其实就是一个循环, 找到一个循环序列即可.

格雷码生成公式
格雷码生成公式: [i] = i^(i>>1) 自己与自己左移一位相异或

i i i >> 1 [i] [i]
0 000 000 000 0
1 001 000 001 1
2 010 001 011 3
3 011 001 010 2
4 100 010 110 6
5 101 010 111 7
6 110 011 101 5
7 111 011 100 4

题解：

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class Solution {
public:
    vector<int> circularPermutation(int n, int start) {
        int size = 1 << n;
        vector<int> arr;

        for(int i = 0; i < size; i++)
        {
            arr.push_back(i ^ (i >> 1));
        }

        vector<int> ans;
        for(int i = 0; i < size; i++)
        {
            if(arr[i] == start)
            {
                for(int j = 0; j < size; j++)
                {
                    ans.push_back(arr[(i+j)%size]);
                }
                return ans;
            }
        }

        return ans;
    }
};

参考资料：
https://www.cnblogs.com/slowbirdoflsh/p/11823498.html