implement-queue-using-stacks

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Algorithms

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>).

Implement the <code>MyQueue</code> class:

<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>

Notes:

<ul>
	<li>You must use only 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&#39;s standard operations.</li>
</ul>

&nbsp;
Example 1:

<pre>
Input
[&quot;MyQueue&quot;, &quot;push&quot;, &quot;push&quot;, &quot;peek&quot;, &quot;pop&quot;, &quot;empty&quot;]
[[], [1], [2], [], [], []]
Output
[null, null, null, 1, 1, false]

Explanation
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>

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= x &lt;= 9</code></li>
	<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>
	<li>All the calls to <code>pop</code> and <code>peek</code> are valid.</li>
</ul>

&nbsp;
Follow-up: Can you implement the queue such that each operation is <a href="https://en.wikipedia.org/wiki/Amortized_analysis" target="_blank">amortized</a> <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.

Implement Queue using Stacks

Implement Stack using Queues

implement-stack-using-queues

用队列实现栈

Stack

Design

Queue

&#35831;&#20320;&#20165;&#20351;&#29992;&#20004;&#20010;&#26632;&#23454;&#29616;&#20808;&#20837;&#20808;&#20986;&#38431;&#21015;&#12290;&#38431;&#21015;&#24212;&#24403;&#25903;&#25345;&#19968;&#33324;&#38431;&#21015;&#25903;&#25345;&#30340;&#25152;&#26377;&#25805;&#20316;&#65288;<code>push</code>&#12289;<code>pop</code>&#12289;<code>peek</code>&#12289;<code>empty</code>&#65289;&#65306;

&#23454;&#29616; <code>MyQueue</code> &#31867;&#65306;

<ul>
	<li><code>void push(int x)</code> &#23558;&#20803;&#32032; x &#25512;&#21040;&#38431;&#21015;&#30340;&#26411;&#23614;</li>
	<li><code>int pop()</code> &#20174;&#38431;&#21015;&#30340;&#24320;&#22836;&#31227;&#38500;&#24182;&#36820;&#22238;&#20803;&#32032;</li>
	<li><code>int peek()</code> &#36820;&#22238;&#38431;&#21015;&#24320;&#22836;&#30340;&#20803;&#32032;</li>
	<li><code>boolean empty()</code> &#22914;&#26524;&#38431;&#21015;&#20026;&#31354;&#65292;&#36820;&#22238; <code>true</code> &#65307;&#21542;&#21017;&#65292;&#36820;&#22238; <code>false</code></li>
</ul>

&#35828;&#26126;&#65306;

<ul>
	<li>&#20320; &#21482;&#33021; &#20351;&#29992;&#26631;&#20934;&#30340;&#26632;&#25805;&#20316; &mdash;&mdash; &#20063;&#23601;&#26159;&#21482;&#26377;&nbsp;<code>push to top</code>,&nbsp;<code>peek/pop from top</code>,&nbsp;<code>size</code>, &#21644;&nbsp;<code>is empty</code>&nbsp;&#25805;&#20316;&#26159;&#21512;&#27861;&#30340;&#12290;</li>
	<li>&#20320;&#25152;&#20351;&#29992;&#30340;&#35821;&#35328;&#20063;&#35768;&#19981;&#25903;&#25345;&#26632;&#12290;&#20320;&#21487;&#20197;&#20351;&#29992; list &#25110;&#32773; deque&#65288;&#21452;&#31471;&#38431;&#21015;&#65289;&#26469;&#27169;&#25311;&#19968;&#20010;&#26632;&#65292;&#21482;&#35201;&#26159;&#26631;&#20934;&#30340;&#26632;&#25805;&#20316;&#21363;&#21487;&#12290;</li>
</ul>

&nbsp;

&#31034;&#20363; 1&#65306;

<pre>
&#36755;&#20837;&#65306;
["MyQueue", "push", "push", "peek", "pop", "empty"]
[[], [1], [2], [], [], []]
&#36755;&#20986;&#65306;
[null, null, null, 1, 1, false]

&#35299;&#37322;&#65306;
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>

&nbsp;

&#25552;&#31034;&#65306;

<ul>
	<li><code>1 &lt;= x &lt;= 9</code></li>
	<li>&#26368;&#22810;&#35843;&#29992; <code>100</code> &#27425; <code>push</code>&#12289;<code>pop</code>&#12289;<code>peek</code> &#21644; <code>empty</code></li>
	<li>&#20551;&#35774;&#25152;&#26377;&#25805;&#20316;&#37117;&#26159;&#26377;&#25928;&#30340; &#65288;&#20363;&#22914;&#65292;&#19968;&#20010;&#31354;&#30340;&#38431;&#21015;&#19981;&#20250;&#35843;&#29992; <code>pop</code> &#25110;&#32773; <code>peek</code> &#25805;&#20316;&#65289;</li>
</ul>

&nbsp;

&#36827;&#38454;&#65306;

<ul>
	<li>&#20320;&#33021;&#21542;&#23454;&#29616;&#27599;&#20010;&#25805;&#20316;&#22343;&#25674;&#26102;&#38388;&#22797;&#26434;&#24230;&#20026; <code>O(1)</code> &#30340;&#38431;&#21015;&#65311;&#25442;&#21477;&#35805;&#35828;&#65292;&#25191;&#34892; <code>n</code> &#20010;&#25805;&#20316;&#30340;&#24635;&#26102;&#38388;&#22797;&#26434;&#24230;&#20026; <code>O(n)</code> &#65292;&#21363;&#20351;&#20854;&#20013;&#19968;&#20010;&#25805;&#20316;&#21487;&#33021;&#33457;&#36153;&#36739;&#38271;&#26102;&#38388;&#12290;</li>
</ul>

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