update

leetcode-cn/originData/words-frequency-lcci.json

@@ -12,7 +12,7 @@
             "translatedContent": "<p>设计一个方法，找出任意指定单词在一本书中的出现频率。</p>\n\n<p>你的实现应该支持如下操作：</p>\n\n<ul>\n\t<li><code>WordsFrequency(book)</code>构造函数，参数为字符串数组构成的一本书</li>\n\t<li><code>get(word)</code>查询指定单词在书中出现的频率</li>\n</ul>\n\n<p><strong>示例：</strong></p>\n\n<pre>WordsFrequency wordsFrequency = new WordsFrequency({&quot;i&quot;, &quot;have&quot;, &quot;an&quot;, &quot;apple&quot;, &quot;he&quot;, &quot;have&quot;, &quot;a&quot;, &quot;pen&quot;});\nwordsFrequency.get(&quot;you&quot;); //返回0，&quot;you&quot;没有出现过\nwordsFrequency.get(&quot;have&quot;); //返回2，&quot;have&quot;出现2次\nwordsFrequency.get(&quot;an&quot;); //返回1\nwordsFrequency.get(&quot;apple&quot;); //返回1\nwordsFrequency.get(&quot;pen&quot;); //返回1\n</pre>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>book[i]</code>中只包含小写字母</li>\n\t<li><code>1 &lt;= book.length &lt;= 100000</code></li>\n\t<li><code>1 &lt;= book[i].length &lt;= 10</code></li>\n\t<li><code>get</code>函数的调用次数不会超过100000</li>\n</ul>\n",
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             "difficulty": "Medium",
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@@ -161,7 +161,7 @@
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             "hints": [
                 "想想这个问题的最佳运行时间是多少。如果你的解法匹配最理想的运行时间，那么你可能无法做的更好了。",
                 "可以使用散列表来优化重复的情况吗？"