[LeetCode] 225. Implement Stack using Queues
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?
用队列实现栈。
请你仅使用两个队列实现一个后入先出（LIFO）的栈，并支持普通栈的全部四种操作（push、top、pop 和 empty）。实现 MyStack 类：
void push(int x) 将元素 x 压入栈顶。
int pop() 移除并返回栈顶元素。
int top() 返回栈顶元素。
boolean empty() 如果栈是空的，返回 true ；否则，返回 false 。注意：
你只能使用队列的基本操作 —— 也就是 push to back、peek/pop from front、size 和 is empty 这些操作。
你所使用的语言也许不支持队列。 你可以使用 list （列表）或者 deque（双端队列）来模拟一个队列 , 只要是标准的队列操作即可。来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/implement-stack-using-queues
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思路
注意这一题的 followup 是问你是否可以只用一个 queue。这一题的 JavaScript 代码会简单很多，因为在 JavaScript 中，队列和栈都是用数组实现的，用法类似 Java 里的 deque。依然解释一下几个函数的实现方式。
push(x) - 将元素推进栈。时间均摊O(n)，因为涉及到会扩大内存的动作。
举个例子，比如此时 queue 中的元素从左到右是 [1, 2, 3]，此时你需要 push 一个新的元素 4，我们要做的是把 4 push 进 queue 的同时将前面所有的元素再拿出来放到 4 的后面，最后 queue 会变成这样
[4, 1, 2, 3]
pop() -- 将栈顶元素弹出。时间O(1)
top() -- 等同于peek，看一下栈顶元素是什么。时间O(1)
empty() -- 判断栈是否为空。时间O(1)
复杂度
同上
代码
Java实现
class MyStack {
Queue<Integer> queue;
public MyStack() {
queue = new LinkedList<>();
}
public void push(int x) {
queue.offer(x);
for (int i = 0; i < queue.size() - 1; i++) {
queue.offer(queue.poll());
}
}
public int pop() {
return queue.poll();
}
public int top() {
return queue.peek();
}
public boolean empty() {
return queue.isEmpty();
}
}
/**
* Your MyStack object will be instantiated and called as such:
* MyStack obj = new MyStack();
* obj.push(x);
* int param_2 = obj.pop();
* int param_3 = obj.top();
* boolean param_4 = obj.empty();
*/
JavaScript实现
/**
* Initialize your data structure here.
*/
var MyStack = function () {
this.stack = [];
};
/**
* Push element x onto stack.
* @param {number} x
* @return {void}
*/
MyStack.prototype.push = function (x) {
this.stack[this.stack.length] = x;
};
/**
* Removes the element on top of the stack and returns that element.
* @return {number}
*/
MyStack.prototype.pop = function () {
let res = this.stack[this.stack.length - 1];
this.stack.length = this.stack.length - 1;
return res;
};
/**
* Get the top element.
* @return {number}
*/
MyStack.prototype.top = function () {
return this.stack[this.stack.length - 1];
};
/**
* Returns whether the stack is empty.
* @return {boolean}
*/
MyStack.prototype.empty = function () {
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()
*/
相关题目
225. Implement Stack using Queues
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