232. Implement Queue using Stacks

🟢   🔖  栈 设计 队列  🔗 LeetCode

题目

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.

题目大意

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

• 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

使用两个栈，将一个栈当作输入栈，用于压入 push 传入的数据；另一个栈当作输出栈，用于 pop 和 peek 操作。

每次 pop 或 peek 时，若输出栈为空则将输入栈的全部数据依次弹出并压入输出栈，这样输出栈从栈顶往栈底的顺序就是队列从队首往队尾的顺序。

代码

class MyQueue {
  constructor() {
    this.pushList = [];
    this.popList = [];
  }

  // @param {number} x
  // @return {void}
  push(x) {
    this.pushList.push(x);
  }

  // @return {number}
  pop() {
    if (this.popList.length === 0) {
      while (this.pushList.length > 0) {
        this.popList.push(this.pushList.pop());
      }
    }
    return this.popList.pop();
  }

  // @return {number}
  peek() {
    if (this.popList.length === 0) {
      while (this.pushList.length > 0) {
        this.popList.push(this.pushList.pop());
      }
    }
    return this.popList[this.popList.length - 1];
  }

  // @return {boolean}
  empty() {
    let len = this.pushList.length + this.popList.length;
    return len === 0;
  }
}

/**
 * Your MyQueue object will be instantiated and called as such:
 * var obj = new MyQueue()
 * obj.push(x)
 * var param_2 = obj.pop()
 * var param_3 = obj.peek()
 * var param_4 = obj.empty()
 */

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