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232. 用栈实现队列


232. 用栈实现队列open in new window

🟢   🔖  设计 队列  🔗 力扣open in new window LeetCodeopen in new window

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 amortizedopen in new window 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(pushpoppeekempty):

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

  • 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 poppeek 操作。

每次 poppeek 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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