Leetcode 3437
Given an integer n, an alternating permutation is a permutation of the first n positive integers such that no two adjacent elements are both odd or both even.
Return all such alternating permutations sorted in lexicographical order.
Example 1:

Input: n = 4
Output: [[1,2,3,4],[1,4,3,2],[2,1,4,3],[2,3,4,1],[3,2,1,4],[3,4,1,2],[4,1,2,3],[4,3,2,1]]

Example 2:

Input: n = 2
Output: [[1,2],[2,1]]

Example 3:

Input: n = 3
Output: [[1,2,3],[3,2,1]]

Constraints:

1 <= n <= 10

此题正常情况是靠模拟就能过，但是由于Leetcode对Ruby进行判定的机制很不友好，基础内存都是206MB往上走，导致n等于10的时候，内存直接爆了。但是Python是没有问题的，如下是相同算法的Ruby和Python实现。
这是Ruby

# @param {Integer} n
# @return {Integer[][]}
def permute(n)
  m = (1..n).to_a.permutation(n).to_a
  ans = []
  m.each do |a|
    flag = false
    a.each_with_index do |a1, i|
      if i < a.length - 1 && (a1 % 2) == (a[i + 1] % 2)
        flag = true
        break
      end
    end
    unless flag
      ans << a
    end
  end
  ans
end

这是Python

from itertools import permutations
class Solution:
    def permute(self, n: int) -> List[List[int]]:
        p = permutations([i for i in range(1,n+1)])
        ans = []
        for p1 in p:
            flag = False
            for i in range(len(p1)):
                if i < len(p1) - 1 and (p1[i]%2) == (p1[i+1]%2):
                    flag = True
                    break
            if not flag:
                ans.append(p1)   
        return ans