225. 用队列实现栈

🟢   🔖  栈 设计 队列  🔗 力扣 LeetCode

题目

Implement a last-in-first-out (LIFO) stack using only two queues. The implemented stack should support all the functions of a normal stack (push, top, pop, and empty).

Implement the MyStack class:

• void push(int x) Pushes element x to the top of the stack.
• int pop() Removes the element on the top of the stack and returns it.
• int top() Returns the element on the top of the stack.
• boolean empty() Returns true if the stack is empty, false otherwise.

Notes:

• You must use only standard operations of a queue, which means that only push to back, peek/pop from front, size and is empty operations are valid.
• Depending on your language, the queue may not be supported natively. You may simulate a queue using a list or deque (double-ended queue) as long as you use only a queue's standard operations.

Example 1:

Input

["MyStack", "push", "push", "top", "pop", "empty"]

[[], [1], [2], [], [], []]

Output

[null, null, null, 2, 2, false]

Explanation

MyStack myStack = new MyStack();

myStack.push(1);

myStack.push(2);

myStack.top(); // return 2

myStack.pop(); // return 2

myStack.empty(); // return False

Constraints:

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

Follow-up: Can you implement the stack using only one queue?

题目大意

题目要求用队列实现一个栈的基本操作：push(x)、pop()、top()、empty()。

解题思路

• 用一个数组来模拟栈，只能使用队列的基本操作，即： push, shift, array[0], array.length
• 每次入栈的时候，将队列内的元素前后颠倒。
• push(x) 的时间复杂度为 O(n)
• pop()、top()、empty() 的时间复杂度为 O(1)

代码

class MyStack {
	constructor() {
		this.stack = [];
	}
	// @param {number} x
	// @return {void}
	push(x) {
		this.stack.push(x);
		for (let i = 0; i < this.stack.length - 1; i++) {
			this.stack.push(this.stack.shift());
		}
	}

	// @return {number}
	pop() {
		return this.stack.shift();
	}

	// @return {number}
	top() {
		return this.stack[0];
	}

	// @return {boolean}
	empty() {
		return this.stack.length === 0;
	}
}

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

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