練習寫過題目的一天~
就算是寫過的題目，過了一陣子還是會有不同的寫法，甚至有些根本忘了怎麼寫

為了不重複貼文，這邊只寫新寫的題目

1137. N-th Tribonacci Number (easy)
      https://leetcode.com/problems/n-th-tribonacci-number/description/
      Submission：https://leetcode.com/problems/n-th-tribonacci-number/submissions/856293958/

The Tribonacci sequence Tn is defined as follows:
T0 = 0, T1 = 1, T2 = 1, and Tn+3 = Tn + Tn+1 + Tn+2 for n >= 0.
Given n, return the value of Tn.
費氏數列是前兩項，這題變前三項，所以寫起來其實感覺一樣。
這邊第一個寫正常解，後面補一個其實我覺得也是正常解啦.....。

class Solution:
    #費氏數列進階款
    def tribonacci(self, n: int) -> int:
        a,b,c,ans = 0,1,1,0
        if n == 0:
            return 0
        elif n <= 2:
            return 1
        while n >= 3:
            ans = a + b + c
            a,b,c = b,c,ans
            n -= 1
        return ans
        
    #n又不大我直接讀表啦，爽(我不做人啦JOJO)
    def tribonacci(self, n: int) -> int:
        nList = [0, 1, 1, 2, 4, 7, 13, 24, 44, 81, 149, 274, 504, 927, 1705, 3136, 5768, 10609, 19513, 35890, 66012, 121415, 223317, 410744, 755476, 1389537, 2555757, 4700770, 8646064, 15902591, 29249425, 53798080, 98950096, 181997601, 334745777, 615693474, 1132436852, 2082876103]
        return nList[n]

938. Range Sum of BST (easy)
     https://leetcode.com/problems/range-sum-of-bst/description/
     Submission：https://leetcode.com/problems/range-sum-of-bst/submissions/856289924/
     Given the root node of a binary search tree and two integers low and high, return the sum of values of all nodes with a value in the inclusive range [low, high].

就讀完整的BST，把low~high之間的數字全部加起來。

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right
class Solution:
    #找出所有在low跟high範圍內的數字加起來
    #這是BST所以可以有一些「排富條款」

    #沒有排富條款
    def rangeSumBST(self, root: Optional[TreeNode], low: int, high: int) -> int:
        def dfs(root):
            if root is None:
                return 0
            temp = root.val if low<=root.val<=high else 0
            return dfs(root.left) + dfs(root.right) + temp
        return dfs(root)

class Solution:
    #找出所有在low跟high範圍內的數字加起來
    #這是BST所以可以有一些「排富條款」

    #有排富條款
    #左邊的child無論數字再大，都不會比他的祖父節點大：
    #似乎不需要，看其他人都沒這樣寫
    #右邊剛好相反
    def rangeSumBST(self, root: Optional[TreeNode], low: int, high: int) -> int:
        def dfs(root):
            if root is None:
                return 0
            if root.val < low:
                return dfs(root.right)
            elif root.val > high:
                return dfs(root.left)
            temp = root.val if low<=root.val<=high else 0
            return dfs(root.left) + dfs(root.right) + temp
        return dfs(root)  

1165. Single-Row Keyboard (easy)
      https://leetcode.com/problems/single-row-keyboard/description/
      Submission：https://leetcode.com/problems/single-row-keyboard/submissions/856274178/

There is a special keyboard with all keys in a single row.

Given a string keyboard of length 26 indicating the layout of the keyboard (indexed from 0 to 25). Initially, your finger is at index 0. To type a character, you have to move your finger to the index of the desired character. The time taken to move your finger from index i to index j is |i - j|.

You want to type a string word. Write a function to calculate how much time it takes to type it with one finger.

class Solution:
    #會有一個串列叫做keyword儲存全部的字母
    #要把word裡所有的字母的索引值找出來，然後計算現在這一個跟下個字母的距離
    def calculateTime(self, keyboard: str, word: str) -> int:
        keyDict = {keyboard[i]:i for i in range(26)}
        ans = temp = keyDict[word[0]]
        for i in word[1:]:
            ans = ans + abs(keyDict[i] - temp) #忘記是算距離，沒有加abs絕對錯
            temp = keyDict[i]
        return ans

以上就是今天的練習感謝大家