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Description
tab: English
<p>Implement a last-in-first-out (LIFO) stack using only two queues. The implemented stack should support all the functions of a normal stack (<code>push</code>, <code>top</code>, <code>pop</code>, and <code>empty</code>).</p>
<p>Implement the <code>MyStack</code> class:</p>
<ul>
<li><code>void push(int x)</code> Pushes element x to the top of the stack.</li>
<li><code>int pop()</code> Removes the element on the top of the stack and returns it.</li>
<li><code>int top()</code> Returns the element on the top of the stack.</li>
<li><code>boolean empty()</code> Returns <code>true</code> if the stack is empty, <code>false</code> otherwise.</li>
</ul>
<p><b>Notes:</b></p>
<ul>
<li>You must use <strong>only</strong> standard operations of a queue, which means that only <code>push to back</code>, <code>peek/pop from front</code>, <code>size</code> and <code>is empty</code> operations are valid.</li>
<li>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.</li>
</ul>
<p> </p>
<p><strong class="example">Example 1:</strong></p>
<pre>
<strong>Input</strong>
["MyStack", "push", "push", "top", "pop", "empty"]
[[], [1], [2], [], [], []]
<strong>Output</strong>
[null, null, null, 2, 2, false]
<strong>Explanation</strong>
MyStack myStack = new MyStack();
myStack.push(1);
myStack.push(2);
myStack.top(); // return 2
myStack.pop(); // return 2
myStack.empty(); // return False
</pre>
<p> </p>
<p><strong>Constraints:</strong></p>
<ul>
<li><code>1 <= x <= 9</code></li>
<li>At most <code>100</code> calls will be made to <code>push</code>, <code>pop</code>, <code>top</code>, and <code>empty</code>.</li>
<li>All the calls to <code>pop</code> and <code>top</code> are valid.</li>
</ul>
<p> </p>
<p><strong>Follow-up:</strong> Can you implement the stack using only one queue?</p>
---
[submissions](https://leetcode.com/problems/implement-stack-using-queues/submissions/) | [solutions](https://leetcode.com/problems/implement-stack-using-queues/solutions/)
tab: 中文
<p>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（<code>push</code>、<code>top</code>、<code>pop</code> 和 <code>empty</code>）。</p>
<p>实现 <code>MyStack</code> 类：</p>
<ul>
<li><code>void push(int x)</code> 将元素 x 压入栈顶。</li>
<li><code>int pop()</code> 移除并返回栈顶元素。</li>
<li><code>int top()</code> 返回栈顶元素。</li>
<li><code>boolean empty()</code> 如果栈是空的，返回 <code>true</code> ；否则，返回 <code>false</code> 。</li>
</ul>
<p> </p>
<p><strong>注意：</strong></p>
<ul>
<li>你只能使用队列的标准操作 —— 也就是 <code>push to back</code>、<code>peek/pop from front</code>、<code>size</code> 和 <code>is empty</code> 这些操作。</li>
<li>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。</li>
</ul>
<p> </p>
<p><strong>示例：</strong></p>
<pre>
<strong>输入：</strong>
["MyStack", "push", "push", "top", "pop", "empty"]
[[], [1], [2], [], [], []]
<strong>输出：</strong>
[null, null, null, 2, 2, false]
<strong>解释：</strong>
MyStack myStack = new MyStack();
myStack.push(1);
myStack.push(2);
myStack.top(); // 返回 2
myStack.pop(); // 返回 2
myStack.empty(); // 返回 False
</pre>
<p> </p>
<p><strong>提示：</strong></p>
<ul>
<li><code>1 <= x <= 9</code></li>
<li>最多调用<code>100</code> 次 <code>push</code>、<code>pop</code>、<code>top</code> 和 <code>empty</code></li>
<li>每次调用 <code>pop</code> 和 <code>top</code> 都保证栈不为空</li>
</ul>
<p> </p>
<p><strong>进阶：</strong>你能否仅用一个队列来实现栈。</p>
---
[提交记录](https://leetcode.cn/problems/implement-stack-using-queues/submissions/) | [题解](https://leetcode.cn/problems/implement-stack-using-queues/solution/)
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