add leetcode problem-cn part2

算法题（国内版）/problem (Chinese)/三合一 [three-in-one-lcci].html

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+<p>三合一。描述如何只用一个数组来实现三个栈。</p>
+
+<p>你应该实现<code>push(stackNum, value)</code>、<code>pop(stackNum)</code>、<code>isEmpty(stackNum)</code>、<code>peek(stackNum)</code>方法。<code>stackNum</code>表示栈下标，<code>value</code>表示压入的值。</p>
+
+<p>构造函数会传入一个<code>stackSize</code>参数，代表每个栈的大小。</p>
+
+<p><strong>示例1:</strong></p>
+
+<pre>
+<strong> 输入</strong>：
+["TripleInOne", "push", "push", "pop", "pop", "pop", "isEmpty"]
+[[1], [0, 1], [0, 2], [0], [0], [0], [0]]
+<strong> 输出</strong>：
+[null, null, null, 1, -1, -1, true]
+<strong>说明</strong>：当栈为空时`pop, peek`返回-1，当栈满时`push`不压入元素。
+</pre>
+
+<p><strong>示例2:</strong></p>
+
+<pre>
+<strong> 输入</strong>：
+["TripleInOne", "push", "push", "push", "pop", "pop", "pop", "peek"]
+[[2], [0, 1], [0, 2], [0, 3], [0], [0], [0], [0]]
+<strong> 输出</strong>：
+[null, null, null, null, 2, 1, -1, -1]
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>0 &lt;= stackNum &lt;= 2</code></li>
+</ul>