题目信息

• 链接: implement-queue-using-stacks

• 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.

  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.

  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.

• 描述:

  • 请你仅使用两个栈实现先入先出队列。队列应当支持一般队列支持的所有操作（push、pop、peek、empty）：
  • 实现 MyQueue 类：
  • void push(int x) 将元素 x 推到队列的末尾
    int pop() 从队列的开头移除并返回元素
    int peek() 返回队列开头的元素
    boolean empty() 如果队列为空，返回 true ；否则，返回 false
    说明：
  • 你 只能 使用标准的栈操作 —— 也就是只有 push to top, peek/pop from top, size, 和 is empty 操作是合法的。
    你所使用的语言也许不支持栈。你可以使用 list 或者 deque（双端队列）来模拟一个栈，只要是标准的栈操作即可。

解题思路

• 思路在于用一个stack1来入队，另一个stack2出队
• 出队时，要让先入队的元素先出，所以要让stack2的顺序和stack1相反。
  • 具体来说就是将stack1元素依次出栈，然后再入栈stack2，这样stack2栈顶就是stack1的栈底，这样stack2出栈的就是队首元素
• 另外就是什么时候将stack1元素倒入stack2？
  • stack2是用来获取队首元素的，所以当pop/peek而stack2为空，stack1不为空时，就将stack1的元素倒入stack2即可

代码

class MyQueue:

    def __init__(self):
        self.stack1 = []
        self.stack2 = []


    def push(self, x: int) -> None:
        self.stack1.append(x)

	# do not need to check empty 
	# since all pop and peek are valid
    def pop(self) -> int:
        self.trans_s1_to_s2()
        return self.stack2.pop()


    def peek(self) -> int:
        self.trans_s1_to_s2()
        return self.stack2[-1]


    def empty(self) -> bool:
        if len(self.stack1) == 0 and len(self.stack2) == 0:
            return True
        else:
            return False

	# if s2 is empty, put all ele in s1 to s2
    def trans_s1_to_s2(self) -> None:
        if len(self.stack2) == 0:
            for _ in range(len(self.stack1)):
                self.stack2.append(self.stack1.pop())



# 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()