###### tags: `阿瑜`

# Day 8: Recap Day [1-7] & Enhance

| Day# | Title | Level |
| ---- | ----- | ----- |
| 1 | [Day 1: Let's start !](https://ithelp.ithome.com.tw/articles/10265412) | None |
| 2 | [Day 2: LeetCode 978. Longest Turbulent Subarray](https://ithelp.ithome.com.tw/articles/10265787) | Medium |
| 3 | [Day 3: LeetCode 995. Minimum Number of K Consecutive Bit Flips](https://ithelp.ithome.com.tw/articles/10267119) | Hard |
| 4 | [Day 4: LeetCode 995. Minimum Number of K Consecutive Bit Flips(v2)](https://ithelp.ithome.com.tw/articles/10267861) | Hard |
| 5 | [Day 5: LeetCode 88. Merge Sorted Array](https://ithelp.ithome.com.tw/articles/10268486) | Easy |
| 6 | [Day 6: LeetCode 54. Spiral Matrix](https://ithelp.ithome.com.tw/articles/10269135) | Medium |
| 7 | [Day 7: LeetCode 485. Max Consecutive Ones](https://ithelp.ithome.com.tw/articles/10269725) | Easy |
### Enhance
- [x] Day 2: LeetCode 978. Longest Turbulent Subarray
- 思路(from Discuss)
- either up or down **begin**

- 🔼 🔽 🔼 🔽 ...
- 🔽 🔼 🔽 🔼 ...
- 繼承上一次(inc/dec)的數量，再加一
- 程式碼

        class Solution:
            def maxTurbulenceSize(self, arr: List[int]) -> int:
                inc = 1
                dec = 1
                res = 1
                for i in range(1,len(arr)):
                    if arr[i] > arr[i-1]:
                        inc = dec+1
                        dec = 1
                    elif arr[i] < arr[i-1]:
                        dec = inc+1
                        inc = 1
                    else:
                        inc = 1
                        dec = 1
                    print("i-inc-dec:",i,inc,dec)    
                    #res = max(inc,dec)
                    res = max([inc,dec,res])
                    print("res:",res)
                return res    

- Reference
- [神人解-Java O(N) time O(1) space](https://leetcode.com/problems/longest-turbulent-subarray/discuss/221935/Java-O(N)-time-O(1)-space)
### What's more?

1. 每月挑戰(2021.09.22)
- [(Medium) LeetCode 1239. Maximum Length of a Concatenated String with Unique Characters](https://leetcode.com/problems/maximum-length-of-a-concatenated-string-with-unique-characters/)
- 題意
- 求unique的**最長**長度
- unique's element 怎麼組成?
Ans: 自己碰其他element所組成
1. 自己中的每個character需unique
2. 碰其他時也需unique
- 思路
- List 中的每個element(set)都是不同組合，在組合中找最長長度
- Python-set() Check 自己中是否有重複element
- 碰前的Check:是否有一樣的character→使用`&`
- 碰(**concate**)所用的運算子`|`
- Reference
- [花花醬-YT圖解](https://youtu.be/N7womGmLXh8)

- 程式碼

        class Solution:
        def maxLength(self, arr: List[str]) -> int:
            A = [set()] # set() => {}
            for a in arr:
                if len(set(a))!=len(a):
                    continue

                for element in A[:]:
                    if set(a)&element:
                        continue
                    A.append(set(a)|element)    
            print(A)            
            maxLen = (-1)            
            for _ in A:
                if len(_)>maxLen:
                    maxLen = len(_)
            return  maxLen               

3. 目的刷
- [(Easy) LeetCode 104. Maximum Depth of Binary Tree](https://leetcode.com/problems/maximum-depth-of-binary-tree/)
- 思路
- BFS-**Queue**
- Initial: Queue 先加root
1. 直到Queue為空
- ans(Heigh)++
3. Iterate(遍歷)每一層節點
- 移除遍歷節點
- 加入遍歷節點的child節點
- We+夥伴
- [soft解(C language)](https://ithelp.ithome.com.tw/articles/10269862)
- [Casey解(Python)](https://ithelp.ithome.com.tw/articles/10270148)
- 其他
- [Iterative Method to find Height of Binary Tree](https://www.google.com/amp/s/www.geeksforgeeks.org/iterative-method-to-find-height-of-binary-tree/amp/)
### What's Next?

> 面試:
> Find Good Job!