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leetcode-cn/problem (Chinese)/计数器 II [counter-ii].html

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+<p>请你写一个函数&nbsp;<code>createCounter</code>. 这个函数接收一个初始的整数值 <code>init</code>&nbsp;&nbsp;并返回一个包含三个函数的对象。</p>
+
+<p>这三个函数是：</p>
+
+<ul>
+	<li><code>increment()</code>&nbsp;将当前值加 1 并返回。</li>
+	<li><code>decrement()</code>&nbsp;将当前值减 1 并返回。</li>
+	<li><code>reset()</code>&nbsp;将当前值设置为 <code>init</code> 并返回。</li>
+</ul>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1：</strong></p>
+
+<pre>
+<b>输入：</b>init = 5, calls = ["increment","reset","decrement"]
+<b>输出：</b>[6,5,4]
+<strong>解释：</strong>
+const counter = createCounter(5);
+counter.increment(); // 6
+counter.reset(); // 5
+counter.decrement(); // 4
+</pre>
+
+<p><strong class="example">示例 2：</strong></p>
+
+<pre>
+<b>输入：</b>init = 0, calls = ["increment","increment","decrement","reset","reset"]
+<b>输出：</b>[1,2,1,0,0]
+<strong>解释：</strong>
+const counter = createCounter(0);
+counter.increment(); // 1
+counter.increment(); // 2
+counter.decrement(); // 1
+counter.reset(); // 0
+counter.reset(); // 0
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>-1000 &lt;= init &lt;= 1000</code></li>
+	<li><code>总调用次数不会超过&nbsp;1000</code></li>
+</ul>