動機

很經典，但沒想過可以用兩個stack

Problem

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]ExplanationMyQueue 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 1myQueue.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.

Sol

這裡是用一個stack去跑

用兩個stack就是一個負責pop的入，一個負責top的read，當需要pop時pop得stack是空的，就從另外一邊搬

class MyQueue:

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

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

    def pop(self) -> int:
        """
        Removes the element from in front of queue and returns that element.
        """
        tmpS = []
        while len(self.stk) > 1:
            tmpS.append(self.stk.pop())
        ret = self.stk.pop()
        while tmpS:
            self.stk.append(tmpS.pop())
        return ret

    def peek(self) -> int:
        """
        Get the front element.
        """
        tmpS = []
        while len(self.stk) > 1:
            tmpS.append(self.stk.pop())
        ret = self.stk.pop()
        tmpS.append(ret)
        while tmpS:
            self.stk.append(tmpS.pop())
        return ret

    def empty(self) -> bool:
        """
        Returns whether the queue is empty.
        """
        return not self.stk