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"question": {
"questionId": "232",
"questionFrontendId": "232",
"categoryTitle": "Algorithms",
"boundTopicId": 1448,
"title": "Implement Queue using Stacks",
"titleSlug": "implement-queue-using-stacks",
"content": "<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>\n\n<p>Implement the <code>MyQueue</code> class:</p>\n\n<ul>\n\t<li><code>void push(int x)</code> Pushes element x to the back of the queue.</li>\n\t<li><code>int pop()</code> Removes the element from the front of the queue and returns it.</li>\n\t<li><code>int peek()</code> Returns the element at the front of the queue.</li>\n\t<li><code>boolean empty()</code> Returns <code>true</code> if the queue is empty, <code>false</code> otherwise.</li>\n</ul>\n\n<p><strong>Notes:</strong></p>\n\n<ul>\n\t<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>\n\t<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&#39;s standard operations.</li>\n</ul>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input</strong>\n[&quot;MyQueue&quot;, &quot;push&quot;, &quot;push&quot;, &quot;peek&quot;, &quot;pop&quot;, &quot;empty&quot;]\n[[], [1], [2], [], [], []]\n<strong>Output</strong>\n[null, null, null, 1, 1, false]\n\n<strong>Explanation</strong>\nMyQueue myQueue = new MyQueue();\nmyQueue.push(1); // queue is: [1]\nmyQueue.push(2); // queue is: [1, 2] (leftmost is front of the queue)\nmyQueue.peek(); // return 1\nmyQueue.pop(); // return 1, queue is [2]\nmyQueue.empty(); // return false\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= x &lt;= 9</code></li>\n\t<li>At most <code>100</code>&nbsp;calls will be made to <code>push</code>, <code>pop</code>, <code>peek</code>, and <code>empty</code>.</li>\n\t<li>All the calls to <code>pop</code> and <code>peek</code> are valid.</li>\n</ul>\n\n<p>&nbsp;</p>\n<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>\n",
"translatedTitle": "用栈实现队列",
"translatedContent": "<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>\n\n<p>实现 <code>MyQueue</code> 类:</p>\n\n<ul>\n\t<li><code>void push(int x)</code> 将元素 x 推到队列的末尾</li>\n\t<li><code>int pop()</code> 从队列的开头移除并返回元素</li>\n\t<li><code>int peek()</code> 返回队列开头的元素</li>\n\t<li><code>boolean empty()</code> 如果队列为空,返回 <code>true</code> ;否则,返回 <code>false</code></li>\n</ul>\n\n<p><strong>说明:</strong></p>\n\n<ul>\n\t<li>你 <strong>只能</strong> 使用标准的栈操作 —— 也就是只有&nbsp;<code>push to top</code>,&nbsp;<code>peek/pop from top</code>,&nbsp;<code>size</code>, 和&nbsp;<code>is empty</code>&nbsp;操作是合法的。</li>\n\t<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>\n</ul>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1</strong></p>\n\n<pre>\n<strong>输入:</strong>\n[\"MyQueue\", \"push\", \"push\", \"peek\", \"pop\", \"empty\"]\n[[], [1], [2], [], [], []]\n<strong>输出:</strong>\n[null, null, null, 1, 1, false]\n\n<strong>解释:</strong>\nMyQueue myQueue = new MyQueue();\nmyQueue.push(1); // queue is: [1]\nmyQueue.push(2); // queue is: [1, 2] (leftmost is front of the queue)\nmyQueue.peek(); // return 1\nmyQueue.pop(); // return 1, queue is [2]\nmyQueue.empty(); // return false\n</pre>\n\n<ul>\n</ul>\n\n<p>&nbsp;</p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= x &lt;= 9</code></li>\n\t<li>最多调用 <code>100</code> 次 <code>push</code>、<code>pop</code>、<code>peek</code> 和 <code>empty</code></li>\n\t<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>\n</ul>\n\n<p>&nbsp;</p>\n\n<p><strong>进阶:</strong></p>\n\n<ul>\n\t<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>\n</ul>\n",
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"code": "type MyQueue struct {\n\n}\n\n\nfunc Constructor() MyQueue {\n\n}\n\n\nfunc (this *MyQueue) Push(x int) {\n\n}\n\n\nfunc (this *MyQueue) Pop() int {\n\n}\n\n\nfunc (this *MyQueue) Peek() int {\n\n}\n\n\nfunc (this *MyQueue) Empty() bool {\n\n}\n\n\n/**\n * Your MyQueue object will be instantiated and called as such:\n * obj := Constructor();\n * obj.Push(x);\n * param_2 := obj.Pop();\n * param_3 := obj.Peek();\n * param_4 := obj.Empty();\n */",
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"code": "class MyQueue\n def initialize()\n\n end\n\n\n=begin\n :type x: Integer\n :rtype: Void\n=end\n def push(x)\n\n end\n\n\n=begin\n :rtype: Integer\n=end\n def pop()\n\n end\n\n\n=begin\n :rtype: Integer\n=end\n def peek()\n\n end\n\n\n=begin\n :rtype: Boolean\n=end\n def empty()\n\n end\n\n\nend\n\n# Your MyQueue object will be instantiated and called as such:\n# obj = MyQueue.new()\n# obj.push(x)\n# param_2 = obj.pop()\n# param_3 = obj.peek()\n# param_4 = obj.empty()",
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"code": "class MyQueue() {\n\n def push(x: Int) {\n\n }\n\n def pop(): Int = {\n\n }\n\n def peek(): Int = {\n\n }\n\n def empty(): Boolean = {\n\n }\n\n}\n\n/**\n * Your MyQueue object will be instantiated and called as such:\n * var obj = new MyQueue()\n * obj.push(x)\n * var param_2 = obj.pop()\n * var param_3 = obj.peek()\n * var param_4 = obj.empty()\n */",
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