Re: [閒聊] 每日leetcode

[dont (dont)]

3217. Delete Nodes From Linked List Present in Array

## 思路

先把nums轉成set
掃linked list時, 如果下一個node的值在set裡面就跳過去

## Code

```python
class Solution:
    def modifiedList(self, nums: List[int], head: Optional[ListNode]) ->
Optional[ListNode]:
        nums = set(nums)

        dummy = ListNode(-1, head)
        curr = dummy
        while curr.next:
            if curr.next.val not in nums:
                curr = curr.next
            else:
                curr.next = curr.next.next

        return dummy.next
```

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640. Solve the Equation

## 思路

parse字串的方式跟前幾天的每日差不多
但要注意x跟0x的case

用兩個變數存 x 跟 value 的值, '='左右邊的+-要反過來

最後檢查無解/無限解 並處理1x, -ax 除gcd

## Code

```python
class Solution:
    def solveEquation(self, equation: str) -> str:

        def get_gcd(x, y):
            while y:
                x, y = y, x % y
            return x

        def parse_val(i, is_left):
            is_x = is_neg = False
            if equation[i] == '+':
                i += 1
            elif equation[i] == '-':
                is_neg = True
                i += 1

            num = 0
            while i < len(equation):
                if equation[i].isdigit():
                    num = num * 10 + int(equation[i])
                    i+=1
                elif equation[i] == 'x':
                    if num == 0 and (i == 0 or equation[i-1] in {'+', '-',
'='}):
                        num = 1
                    is_x = True
                    i+=1
                else:
                    break

            if is_neg:
                num = -num

            if is_x:
                res[0] += num * (1 if is_left else -1)
            else:
                res[1] += num * (-1 if is_left else 1)

            return i

        i, n = 0, len(equation)
        res = [0, 0]
        is_left = True
        while i < n:
            i = parse_val(i, is_left)
            if i < n and equation[i] == '=':
                is_left = False
                i += 1

        if res[0] == 0:
            if res[1] == 0:
                return 'Infinite solutions'
            return 'No solution'

        gcd = get_gcd(res[0], res[1])
        res[0] //= gcd
        res[1] //= gcd

        if res[0] < 0:
            res[0], res[1] = -res[0], -res[1]

        if res[0] == 1:
            return f'x={res[1]}'

        return f'{res[0]}x={res[1]}'
```

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