You are given a?0-indexed?array?nums?containing?n?integers.

At each second, you perform the following operation on the array:

• For every index?i?in the range?[0, n - 1], replace?nums[i]?with either?nums[i],?nums[(i - 1 + n) % n], or?nums[(i + 1) % n].

Note?that all the elements get replaced simultaneously.

Return?the?minimum?number of seconds needed to make all elements in the array?nums?equal.

?

Example 1:

Input: nums = [1,2,1,2]

Output: 1

Explanation:?

We can equalize the array in 1 second in the following way: - At 1st second, replace values at each index with [nums[3],nums[1],nums[3],nums[3]]. After replacement, nums = [2,2,2,2]. It can be proven that 1 second is the minimum amount of seconds needed for equalizing the array.

Example 2:

Input: nums = [2,1,3,3,2]

Output: 2

Explanation:

We can equalize the array in 2 seconds in the following way: - At 1st second, replace values at each index with [nums[0],nums[2],nums[2],nums[2],nums[3]]. After replacement, nums = [2,3,3,3,3]. - At 2nd second, replace values at each index with [nums[1],nums[1],nums[2],nums[3],nums[4]]. After replacement, nums = [3,3,3,3,3]. It can be proven that 2 seconds is the minimum amount of seconds needed for equalizing the array.

Example 3:

Input: nums = [5,5,5,5]

Output: 0

Explanation:?

We don't need to perform any operations as all elements in the initial array are the same.

?

Constraints:

• 1 <= n == nums.length <= 105

• 1 <= nums[i] <= 109

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給你一個下標從?0?開始長度為?n?的數(shù)組?nums?。

每一秒，你可以對數(shù)組執(zhí)行以下操作：

• 對于范圍在?[0, n - 1]?內(nèi)的每一個下標?i?，將?nums[i]?替換成?nums[i]?，nums[(i - 1 + n) % n]?或者?nums[(i + 1) % n]?三者之一。

注意，所有元素會被同時替換。

請你返回將數(shù)組?nums?中所有元素變成相等元素所需要的?最少?秒數(shù)。

?

示例 1：

輸入：nums = [1,2,1,2]

輸出：1

解釋：我們可以在 1 秒內(nèi)將數(shù)組變成相等元素： - 第 1 秒，將每個位置的元素分別變?yōu)?[nums[3],nums[1],nums[3],nums[3]] 。變化后，nums = [2,2,2,2] 。 1 秒是將數(shù)組變成相等元素所需要的最少秒數(shù)。

示例 2：

輸入：nums = [2,1,3,3,2]

輸出：2

解釋：我們可以在 2 秒內(nèi)將數(shù)組變成相等元素： - 第 1 秒，將每個位置的元素分別變?yōu)?[nums[0],nums[2],nums[2],nums[2],nums[3]] 。變化后，nums = [2,3,3,3,3] 。 - 第 2 秒，將每個位置的元素分別變?yōu)?[nums[1],nums[1],nums[2],nums[3],nums[4]] 。變化后，nums = [3,3,3,3,3] 。 2 秒是將數(shù)組變成相等元素所需要的最少秒數(shù)。

示例 3：

輸入：nums = [5,5,5,5]

輸出：0

解釋：不需要執(zhí)行任何操作，因為一開始數(shù)組中的元素已經(jīng)全部相等。

?

Runtime:?77 ms, faster than?49.12%?of?Java?online submissions for?Minimum Seconds to Equalize a Circular Array.

Memory Usage:?90.3 MB, less than?6.80%?of?Java?online submissions for?Minimum Seconds to Equalize a Circular Array.

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如果數(shù)組中沒有相同的數(shù)字，那么肯定話費的時間是最長的也就是n/2了，既然可以改變左右2個相鄰的數(shù)字，那么我們將相同元素的索引值放到一個鏈表里面，注意最后加一個n，這樣就是環(huán)形鏈表了，

然后依次去判斷2個index的最值即可，

最后跟n/2取最小值即可；

下面是代碼：