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leetcode-cn/problem (Chinese)/堆盘子 [stack-of-plates-lcci].html
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<p>当某个栈为空时，应当删除该栈。当栈中没有元素或不存在该栈时，<code>pop</code>，<code>popAt</code>&nbsp;应返回 -1.</p>
<p><strong>示例1:</strong></p>
<p><strong>示例 1：</strong></p>
<pre><strong> 输入</strong>：
[&quot;StackOfPlates&quot;, &quot;push&quot;, &quot;push&quot;, &quot;popAt&quot;, &quot;pop&quot;, &quot;pop&quot;]
<pre>
<strong> 输入</strong>：
["StackOfPlates", "push", "push", "popAt", "pop", "pop"]
[[1], [1], [2], [1], [], []]
<strong> 输出</strong>：
[null, null, null, 2, 1, -1]
</pre>
<p><strong>示例2:</strong></p>
<p><strong>示例 2：</strong></p>
<pre><strong> 输入</strong>：
[&quot;StackOfPlates&quot;, &quot;push&quot;, &quot;push&quot;, &quot;push&quot;, &quot;popAt&quot;, &quot;popAt&quot;, &quot;popAt&quot;]
<pre>
<strong> 输入</strong>：
["StackOfPlates", "push", "push", "push", "popAt", "popAt", "popAt"]
[[2], [1], [2], [3], [0], [0], [0]]
<strong> 输出</strong>：
[null, null, null, null, 2, 1, 3]