【leetcode】1243. Array Transformation

题目如下：

Given an initial array arr, every day you produce a new array using the array of the previous day.

On the i-th day, you do the following operations on the array of day i-1 to produce the array of day i:

1. If an element is smaller than both its left neighbor and its right neighbor, then this element is incremented.
2. If an element is bigger than both its left neighbor and its right neighbor, then this element is decremented.
3. The first and last elements never change.

After some days, the array does not change. Return that final array.

Example 1:

Input: arr = [6,2,3,4]
Output: [6,3,3,4]
Explanation: 
On the first day, the array is changed from [6,2,3,4] to [6,3,3,4].
No more operations can be done to this array.

Example 2:

Input: arr = [1,6,3,4,3,5]
Output: [1,4,4,4,4,5]
Explanation: 
On the first day, the array is changed from [1,6,3,4,3,5] to [1,5,4,3,4,5].
On the second day, the array is changed from [1,5,4,3,4,5] to [1,4,4,4,4,5].
No more operations can be done to this array.

Constraints:

• 1 <= arr.length <= 100
• 1 <= arr[i] <= 100

解题思路：题目不难，注意每次变换前先备份arr，再根据备份的arr的值修改本身的arr。

代码如下：

class Solution(object):
    def transformArray(self, arr):
        """
        :type arr: List[int]
        :rtype: List[int]
        """
        while True:
            arr_bak = arr[::]
            for i in range(1,len(arr_bak)-1):
                if arr_bak[i] > arr_bak[i-1] and arr_bak[i] > arr_bak[i+1]:
                    arr[i] -= 1
                elif arr_bak[i] < arr_bak[i-1] and arr_bak[i] < arr_bak[i+1]:
                    arr[i] += 1
            if arr_bak == arr:
                break
        return arr