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Description
tab: English
<p>Implement a first in first out (FIFO) queue using only two stacks. The implemented queue should support all the functions of a normal queue (<code>push</code>, <code>peek</code>, <code>pop</code>, and <code>empty</code>).</p>
<p>Implement the <code>MyQueue</code> class:</p>
<ul>
<li><code>void push(int x)</code> Pushes element x to the back of the queue.</li>
<li><code>int pop()</code> Removes the element from the front of the queue and returns it.</li>
<li><code>int peek()</code> Returns the element at the front of the queue.</li>
<li><code>boolean empty()</code> Returns <code>true</code> if the queue is empty, <code>false</code> otherwise.</li>
</ul>
<p><strong>Notes:</strong></p>
<ul>
<li>You must use <strong>only</strong> standard operations of a stack, which means only <code>push to top</code>, <code>peek/pop from top</code>, <code>size</code>, and <code>is empty</code> operations are valid.</li>
<li>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.</li>
</ul>
<p> </p>
<p><strong class="example">Example 1:</strong></p>
<pre>
<strong>Input</strong>
["MyQueue", "push", "push", "peek", "pop", "empty"]
[[], [1], [2], [], [], []]
<strong>Output</strong>
[null, null, null, 1, 1, false]
<strong>Explanation</strong>
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
</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>peek</code>, and <code>empty</code>.</li>
<li>All the calls to <code>pop</code> and <code>peek</code> are valid.</li>
</ul>
<p> </p>
<p><strong>Follow-up:</strong> Can you implement the queue such that each operation is <strong><a href="https://en.wikipedia.org/wiki/Amortized_analysis" target="_blank">amortized</a></strong> <code>O(1)</code> time complexity? In other words, performing <code>n</code> operations will take overall <code>O(n)</code> time even if one of those operations may take longer.</p>
---
[submissions](https://leetcode.com/problems/implement-queue-using-stacks/submissions/) | [solutions](https://leetcode.com/problems/implement-queue-using-stacks/solutions/)
tab: 中文
<p>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（<code>push</code>、<code>pop</code>、<code>peek</code>、<code>empty</code>）：</p>
<p>实现 <code>MyQueue</code> 类：</p>
<ul>
<li><code>void push(int x)</code> 将元素 x 推到队列的末尾</li>
<li><code>int pop()</code> 从队列的开头移除并返回元素</li>
<li><code>int peek()</code> 返回队列开头的元素</li>
<li><code>boolean empty()</code> 如果队列为空，返回 <code>true</code> ；否则，返回 <code>false</code></li>
</ul>
<p><strong>说明：</strong></p>
<ul>
<li>你 <strong>只能</strong> 使用标准的栈操作 —— 也就是只有 <code>push to top</code>, <code>peek/pop from top</code>, <code>size</code>, 和 <code>is empty</code> 操作是合法的。</li>
<li>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。</li>
</ul>
<p> </p>
<p><strong>示例 1：</strong></p>
<pre>
<strong>输入：</strong>
["MyQueue", "push", "push", "peek", "pop", "empty"]
[[], [1], [2], [], [], []]
<strong>输出：</strong>
[null, null, null, 1, 1, false]
<strong>解释：</strong>
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
</pre>
<ul>
</ul>
<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>peek</code> 和 <code>empty</code></li>
<li>假设所有操作都是有效的 （例如，一个空的队列不会调用 <code>pop</code> 或者 <code>peek</code> 操作）</li>
</ul>
<p> </p>
<p><strong>进阶：</strong></p>
<ul>
<li>你能否实现每个操作均摊时间复杂度为 <code>O(1)</code> 的队列？换句话说，执行 <code>n</code> 个操作的总时间复杂度为 <code>O(n)</code> ，即使其中一个操作可能花费较长时间。</li>
</ul>
---
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