Implement a first in first out (FIFO) queue using only two stacks. The implemented queue should support all the functions of a normal queue (push, peek, pop, and empty).

Implement the MyQueue class:

  • void push(int x) Pushes element x to the back of the queue.
  • int pop() Removes the element from the front of the queue and returns it.
  • int peek() Returns the element at the front of the queue.
  • boolean empty() Returns true if the queue is empty, false otherwise.

Notes:

  • You must use only standard operations of a stack, which means only push to top, peek/pop from top, size, and is empty operations are valid.
  • Depending on your language, the stack may not be supported natively. You may simulate a stack using a list or deque (double-ended queue) as long as you use only a stack’s standard operations.
    Follow-up: Can you implement the queue such that each operation is amortized O(1) time complexity? In other words, performing n operations will take overall O(n) time even if one of those operations may take longer.

Example 1:
Input
["MyQueue", "push", "push", "peek", "pop", "empty"]
[[], [1], [2], [], [], []]
Output
[null, null, null, 1, 1, false]

Explanation
MyQueue myQueue = new MyQueue();
myQueue.push(1); // queue is: [1]
myQueue.push(2); // queue is: [1, 2] (leftmost is front of the queue)
myQueue.peek(); // return 1
myQueue.pop(); // return 1, queue is [2]
myQueue.empty(); // return false

Constraints:

  • 1 <= x <= 9
  • At most 100 calls will be made to push, pop, peek, and empty.
  • All the calls to pop and peek are valid.

Solution in python:

class MyQueue:

    def __init__(self):
        """
        Initialize your data structure here.
        """
        self.stack1 = []
        self.stack2 = []

    def push(self, x: int) -> None:
        """
        Push element x to the back of queue.
        """
        self.stack1.append(x)

    def pop(self) -> int:
        """
        Removes the element from in front of queue and returns that element.
        """
        size = len(self.stack1)
        while size != 0:
            self.stack2.append(self.stack1.pop())
            size -= 1
        temp = self.stack2.pop()
        while len(self.stack2) != 0:
            self.stack1.append(self.stack2.pop())
        return temp  

    def peek(self) -> int:
        """
        Get the front element.
        """
        size = len(self.stack1)
        while size != 0:
            self.stack2.append(self.stack1.pop())
            size -= 1
        temp = self.stack2.pop()
        self.stack1.append(temp)
        while len(self.stack2) != 0:
            self.stack1.append(self.stack2.pop())
        return temp

    def empty(self) -> bool:
        """
        Returns whether the queue is empty.
        """
        return len(self.stack1) == 0

# Your MyQueue object will be instantiated and called as such:
# obj = MyQueue()
# obj.push(x)
# param_2 = obj.pop()
# param_3 = obj.peek()
# param_4 = obj.empty()
最后修改日期: 2021年1月21日

留言

撰写回覆或留言

发布留言必须填写的电子邮件地址不会公开。