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

Recap

Day#	Title	Level
1	Day 1: Let's start !	None
2	Day 2: LeetCode 978. Longest Turbulent Subarray	Medium
3	Day 3: LeetCode 995. Minimum Number of K Consecutive Bit Flips	Hard
4	Day 4: LeetCode 995. Minimum Number of K Consecutive Bit Flips(v2)	Hard
5	Day 5: LeetCode 88. Merge Sorted Array	Easy
6	Day 6: LeetCode 54. Spiral Matrix	Medium
7	Day 7: LeetCode 485. Max Consecutive Ones	Easy

Enhance

[Ｖ] Day 2: LeetCode 978. Longest Turbulent Subarray

思路(from Discuss)
- either up or down begin
  - 上下上下 ...
  - 下上下上 ...
- 继承上一次(inc/dec)的数量，再加一

程序码

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

What's more?

每月挑战(2021.09.22)
- (Medium) LeetCode 1239. Maximum Length of a Concatenated String with Unique Characters
- 题意
  - 求unique的最长长度
  - unique's element 怎麽组成?
    Ans: 自己碰其他element所组成
    1. 自己中的每个character需unique
    2. 碰其他时也需unique
- 思路
  - List 中的每个element(set)都是不同组合，在组合中找最长长度
  - Python-set() Check 自己中是否有重复element
  - 碰前的Check:是否有一样的character→使用&
  - 碰(concate)所用的运算子|
  - Reference
    - 花花酱-YT图解
- 程序码
```
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               
```
目的刷
- (Easy) LeetCode 104. Maximum Depth of Binary Tree
- 思路
  - BFS-Queue
    - Initial: Queue 先加root
    1. 直到Queue为空
      - ans(Heigh)++
    2. Iterate(遍历)每一层节点
      - 移除遍历节点
      - 加入遍历节点的child节点
- We+夥伴
  - soft解(C language)
  - Casey解(Python)
- 其他
  - Iterative Method to find Height of Binary Tree