You are given a 0-indexed array nums of distinct integers. You want to rearrange the elements in the array such that every element in the rearranged array is not equal to the average of its neighbors.
More formally, the rearranged array should have the property such that for every i in the range 1 <= i < nums.length - 1, (nums[i-1] + nums[i+1]) / 2 is not equal to nums[i].
Return any rearrangement of nums that meets the requirements.
\n
Example 1:
\n\n\nInput: nums = [1,2,3,4,5]\nOutput: [1,2,4,5,3]\nExplanation:\nWhen i=1, nums[i] = 2, and the average of its neighbors is (1+4) / 2 = 2.5.\nWhen i=2, nums[i] = 4, and the average of its neighbors is (2+5) / 2 = 3.5.\nWhen i=3, nums[i] = 5, and the average of its neighbors is (4+3) / 2 = 3.5.\n\n\n
Example 2:
\n\n\nInput: nums = [6,2,0,9,7]\nOutput: [9,7,6,2,0]\nExplanation:\nWhen i=1, nums[i] = 7, and the average of its neighbors is (9+6) / 2 = 7.5.\nWhen i=2, nums[i] = 6, and the average of its neighbors is (7+2) / 2 = 4.5.\nWhen i=3, nums[i] = 2, and the average of its neighbors is (6+0) / 2 = 3.\n\n\n
\n
Constraints:
\n\n3 <= nums.length <= 1050 <= nums[i] <= 105给你一个 下标从 0 开始 的数组 nums ，数组由若干 互不相同的 整数组成。你打算重新排列数组中的元素以满足：重排后，数组中的每个元素都 不等于 其两侧相邻元素的 平均值 。
更公式化的说法是，重新排列的数组应当满足这一属性：对于范围 1 <= i < nums.length - 1 中的每个 i ，(nums[i-1] + nums[i+1]) / 2 不等于 nums[i] 均成立 。
返回满足题意的任一重排结果。
\n\n\n\n
示例 1：
\n\n输入：nums = [1,2,3,4,5]\n输出：[1,2,4,5,3]\n解释：\ni=1, nums[i] = 2, 两相邻元素平均值为 (1+4) / 2 = 2.5\ni=2, nums[i] = 4, 两相邻元素平均值为 (2+5) / 2 = 3.5\ni=3, nums[i] = 5, 两相邻元素平均值为 (4+3) / 2 = 3.5\n\n\n
示例 2：
\n\n输入：nums = [6,2,0,9,7]\n输出：[9,7,6,2,0]\n解释：\ni=1, nums[i] = 7, 两相邻元素平均值为 (9+6) / 2 = 7.5\ni=2, nums[i] = 6, 两相邻元素平均值为 (7+2) / 2 = 4.5\ni=3, nums[i] = 2, 两相邻元素平均值为 (6+0) / 2 = 3\n\n\n
\n\n
提示：
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