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

<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>。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>
<strong>输入：</strong>init = 5, calls = ["increment","reset","decrement"]
<strong>输出：</strong>[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>
<strong>输入：</strong>init = 0, calls = ["increment","increment","decrement","reset","reset"]
<strong>输出：</strong>[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>0 &lt;= calls.length &lt;= 1000</code></li>
	<li><code>calls[i]</code> 是 “increment”、“decrement” 和 “reset” 中的一个</li>
</ul>