LeetCode 上极少见的工程级滑窗实现

我Pythonic风格代码，看来有所长进，这源于我Java深厚的OOP功底。

✅ 最终评价

AI评价: 这段代码已经是 LeetCode 上极少见的工程级滑窗实现。

from dataclasses import dataclass
from heapq import heappush, heappop

@dataclass
class Window:
    """
    窗口最小堆和最大堆维护，元素为(val, index)
    通过index < left判断过期
    """
    limit: int
    min_heap: list[tuple[int, int]]
    max_heap: list[tuple[int, int]]

    def push(self, *, val: int, idx: int):
        """
        入窗
        """
        heappush(self.min_heap, (val, idx))
        heappush(self.max_heap, (-val, idx))

    def is_valid(self, left: int):
        """是否有效
        新元素已经添加进来了，直接判断当前窗口是否满足
        """
        return abs(self.min_heap[0][0] + self.max_heap[0][0]) <= self.limit

    def clean(self, left: int):
        """出窗
        清除过期的元素
        """
        while self.min_heap[0][1] < left:
            heappop(self.min_heap)
        while self.max_heap[0][1] < left:
            heappop(self.max_heap)


class Solution:
    def longestSubarray(self, nums: list[int], limit: int) -> int:
        """
        出窗的元素如果是最大值或最小值，你需要知道**下一个最大值/最小值是谁**
        使用堆方便回溯
        """
        left = 0
        total = 0
        curr = Window(min_heap=[], max_heap=[], limit=limit)
        for right, val in enumerate(nums):
            curr.push(val=val, idx=right)
            while not curr.is_valid(left):
                left += 1
                curr.clean(left)
            total = max(total, right - left + 1)
        return total