【Leetcode Daily】1670设计前中后队列

题目浅析

• 想查看原题可以点击题目链接。

• 简单地说，就是设计一个队列，支持在这个队列的前中后三个地方进行入队和出队的操作。

思路分享

• 借鉴灵山的不错的思路，用两个队列来模拟，规定，两个队列大小相同时，优先入队右侧，所以出队时也优先右侧的左端。

  https://leetcode.cn/problems/design-front-middle-back-queue/solutions/2544784/tu-jie-liang-ge-shuang-duan-dui-lie-jian-43pr/

代码解答（强烈建议自行解答后再看）

• 参考题解

class FrontMiddleBackQueue:
    def __init__(self):
        self.left = deque()
        self.right = deque()

    def balance(self):
        if len(self.left) > len(self.right):
            self.right.appendleft(self.left.pop())
        elif len(self.right) > len(self.left)+1:
            self.left.append(self.right.popleft())

    def pushFront(self, val: int) -> None:
        self.left.appendleft(val)
        self.balance()

    def pushMiddle(self, val: int) -> None:
        if len(self.left) < len(self.right):
            self.left.append(val)
        else:
            self.right.appendleft(val)

    def pushBack(self, val: int) -> None:
        self.right.append(val)
        self.balance()

    def popFront(self) -> int:
        if not self.right:  # 整个队列为空
            return -1
        val = self.left.popleft() if self.left else self.right.popleft()
        self.balance()
        return val

    def popMiddle(self) -> int:
        if not self.right:  # 整个队列为空
            return -1
        if len(self.left) == len(self.right):
            return self.left.pop()
        return self.right.popleft()

    def popBack(self) -> int:
        if not self.right:  # 整个队列为空
            return -1
        val = self.right.pop()
        self.balance()
        return val

# Your FrontMiddleBackQueue object will be instantiated and called as such:
# obj = FrontMiddleBackQueue()
# obj.pushFront(val)
# obj.pushMiddle(val)
# obj.pushBack(val)
# param_4 = obj.popFront()
# param_5 = obj.popMiddle()
# param_6 = obj.popBack()