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小墨
2022-05-02 23:44:12 +08:00
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leetcode-cn/originData/design-browser-history.json
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@@ -12,7 +12,7 @@
"translatedContent": "<p>你有一个只支持单个标签页的 <strong>浏览器</strong>&nbsp;，最开始你浏览的网页是&nbsp;<code>homepage</code>&nbsp;，你可以访问其他的网站&nbsp;<code>url</code>&nbsp;，也可以在浏览历史中后退&nbsp;<code>steps</code>&nbsp;步或前进&nbsp;<code>steps</code>&nbsp;步。</p>\n\n<p>请你实现&nbsp;<code>BrowserHistory</code> 类：</p>\n\n<ul>\n\t<li><code>BrowserHistory(string homepage)</code>&nbsp;，用&nbsp;<code>homepage</code>&nbsp;初始化浏览器类。</li>\n\t<li><code>void visit(string url)</code>&nbsp;从当前页跳转访问 <code>url</code> 对应的页面&nbsp;&nbsp;。执行此操作会把浏览历史前进的记录全部删除。</li>\n\t<li><code>string back(int steps)</code>&nbsp;在浏览历史中后退&nbsp;<code>steps</code>&nbsp;步。如果你只能在浏览历史中后退至多&nbsp;<code>x</code> 步且&nbsp;<code>steps &gt; x</code>&nbsp;，那么你只后退&nbsp;<code>x</code>&nbsp;步。请返回后退 <strong>至多</strong> <code>steps</code>&nbsp;步以后的&nbsp;<code>url</code>&nbsp;。</li>\n\t<li><code>string forward(int steps)</code>&nbsp;在浏览历史中前进&nbsp;<code>steps</code>&nbsp;步。如果你只能在浏览历史中前进至多&nbsp;<code>x</code>&nbsp;步且&nbsp;<code>steps &gt; x</code>&nbsp;，那么你只前进 <code>x</code>&nbsp;步。请返回前进&nbsp;<strong>至多</strong>&nbsp;<code>steps</code>步以后的 <code>url</code>&nbsp;。</li>\n</ul>\n\n<p>&nbsp;</p>\n\n<p><strong>示例：</strong></p>\n\n<pre><strong>输入：</strong>\n[&quot;BrowserHistory&quot;,&quot;visit&quot;,&quot;visit&quot;,&quot;visit&quot;,&quot;back&quot;,&quot;back&quot;,&quot;forward&quot;,&quot;visit&quot;,&quot;forward&quot;,&quot;back&quot;,&quot;back&quot;]\n[[&quot;leetcode.com&quot;],[&quot;google.com&quot;],[&quot;facebook.com&quot;],[&quot;youtube.com&quot;],[1],[1],[1],[&quot;linkedin.com&quot;],[2],[2],[7]]\n<strong>输出：</strong>\n[null,null,null,null,&quot;facebook.com&quot;,&quot;google.com&quot;,&quot;facebook.com&quot;,null,&quot;linkedin.com&quot;,&quot;google.com&quot;,&quot;leetcode.com&quot;]\n\n<strong>解释：</strong>\nBrowserHistory browserHistory = new BrowserHistory(&quot;leetcode.com&quot;);\nbrowserHistory.visit(&quot;google.com&quot;); // 你原本在浏览 &quot;leetcode.com&quot; 。访问 &quot;google.com&quot;\nbrowserHistory.visit(&quot;facebook.com&quot;); // 你原本在浏览 &quot;google.com&quot; 。访问 &quot;facebook.com&quot;\nbrowserHistory.visit(&quot;youtube.com&quot;); // 你原本在浏览 &quot;facebook.com&quot; 。访问 &quot;youtube.com&quot;\nbrowserHistory.back(1); // 你原本在浏览 &quot;youtube.com&quot; ，后退到 &quot;facebook.com&quot; 并返回 &quot;facebook.com&quot;\nbrowserHistory.back(1); // 你原本在浏览 &quot;facebook.com&quot; ，后退到 &quot;google.com&quot; 并返回 &quot;google.com&quot;\nbrowserHistory.forward(1); // 你原本在浏览 &quot;google.com&quot; ，前进到 &quot;facebook.com&quot; 并返回 &quot;facebook.com&quot;\nbrowserHistory.visit(&quot;linkedin.com&quot;); // 你原本在浏览 &quot;facebook.com&quot; 。 访问 &quot;linkedin.com&quot;\nbrowserHistory.forward(2); // 你原本在浏览 &quot;linkedin.com&quot; ，你无法前进任何步数。\nbrowserHistory.back(2); // 你原本在浏览 &quot;linkedin.com&quot; ，后退两步依次先到 &quot;facebook.com&quot; ，然后到 &quot;google.com&quot; ，并返回 &quot;google.com&quot;\nbrowserHistory.back(7); // 你原本在浏览 &quot;google.com&quot;， 你只能后退一步到 &quot;leetcode.com&quot; ，并返回 &quot;leetcode.com&quot;\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= homepage.length &lt;= 20</code></li>\n\t<li><code>1 &lt;= url.length &lt;= 20</code></li>\n\t<li><code>1 &lt;= steps &lt;= 100</code></li>\n\t<li><code>homepage</code> 和&nbsp;<code>url</code>&nbsp;都只包含&nbsp;&#39;.&#39; 或者小写英文字母。</li>\n\t<li>最多调用&nbsp;<code>5000</code>&nbsp;次&nbsp;<code>visit</code>，&nbsp;<code>back</code>&nbsp;和&nbsp;<code>forward</code>&nbsp;函数。</li>\n</ul>\n",
"isPaidOnly": false,
"difficulty": "Medium",
"likes": 45,
"likes": 47,
"dislikes": 0,
"isLiked": null,
"similarQuestions": "[]",
@@ -167,7 +167,7 @@
"__typename": "CodeSnippetNode"
}
],
"stats": "{\"totalAccepted\": \"10.6K\", \"totalSubmission\": \"17.2K\", \"totalAcceptedRaw\": 10648, \"totalSubmissionRaw\": 17215, \"acRate\": \"61.9%\"}",
"stats": "{\"totalAccepted\": \"11.1K\", \"totalSubmission\": \"17.9K\", \"totalAcceptedRaw\": 11082, \"totalSubmissionRaw\": 17929, \"acRate\": \"61.8%\"}",
"hints": [
"Use two stack one for back history and one for forward history and simulate the functions.",
"Can you do faster by using different data structure ?"