LeetCode 5096. Array transform

Address  https://leetcode-cn.com/contest/biweekly-contest-12/problems/array-transformation/

First of all, give you an initial array arr. Then, every day you have to generate a new array of arrays based on the previous day.

The resulting array on day i, i-1 is the first day of the array obtained by the following operations you:

If an element is less than its left and right neighbors, then the element is incremented by one.
If an element is greater than its left and right neighbors, then the element 1 is decremented.
The first and last elements never change.
After a while, you'll find an array will not change, return an array finally obtained.

Example 1 : 

Input: [ 6 , 2 , 3 , 4 ] 
Output: [ 6 , 3 , 3 , 4 ] 
Explanation: 
The first day, the array [ 6 , 2 , 3 , 4 ] to [ 6 , 3 , 3 , 4 ]. 
No longer the array more. 
Example 2 : 

Input: [ 1 , 6 , 3 , 4 , 3 , 5 ] 
Output: [1 , 4 , 4 , 4 , 4 , 5 ] 
Explanation: 
The first day, the array [ 1 , 6 , 3 , 4 , 3 , 5 ] to [ 1 , 5 , 4 , 3 , 4 , 5 ]. 
The next day, the array [ 1 , 5 , 4 , 3 , 4 , 5 ] to [ 1 , 4 , 4 , 4 ,4 , 5 ]. 
No longer the array more.

Algorithm 1
This question is just barely make out the optimization of the location, there are many
full simulation title meaning
each cycle opened a pig and ask array as long array for each element in turn to detect if they meet the meaning of the questions in the newly opened array +1 or -1 record
finally opened a new array added to the original array
if the array to open all zeros so that no changes can be out of the loop

 1 class Solution {
 2 public:
 3     vector<int> transformArray(vector<int>& arr) {
 4         if(arr.size() == 1 || arr.size() == 2) return arr;
 5 
 6         while(1){
 7             vector<int> addvec(arr.size(),0);
 8             for(int i = 1;i < arr.size()-1;i++){
 9                 if(arr[i] > arr[i-1] && arr[i]>arr[i+1]) addvec[i] = -1;
10                 else if(arr[i] < arr[i-1] && arr[i]<arr[i+1]) addvec[i] = 1;
11             }
12             int isAllZero = true;
13             for(int i = 0; i < addvec.size();i++){
14                 arr[i] += addvec[i];
15                 if(addvec[i] != 0) isAllZero =false;
16             } 
17             if(isAllZero) break;
18         }
19 
20 
21         return arr;
22     }
23 };

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