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            "categoryTitle": "Algorithms",
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            "title": "Backspace String Compare",
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            "content": "<p>Given two strings <code>s</code> and <code>t</code>, return <code>true</code> <em>if they are equal when both are typed into empty text editors</em>. <code>&#39;#&#39;</code> means a backspace character.</p>\n\n<p>Note that after backspacing an empty text, the text will continue empty.</p>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;ab#c&quot;, t = &quot;ad#c&quot;\n<strong>Output:</strong> true\n<strong>Explanation:</strong> Both s and t become &quot;ac&quot;.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;ab##&quot;, t = &quot;c#d#&quot;\n<strong>Output:</strong> true\n<strong>Explanation:</strong> Both s and t become &quot;&quot;.\n</pre>\n\n<p><strong>Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;a#c&quot;, t = &quot;b&quot;\n<strong>Output:</strong> false\n<strong>Explanation:</strong> s becomes &quot;c&quot; while t becomes &quot;b&quot;.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code><span>1 &lt;= s.length, t.length &lt;= 200</span></code></li>\n\t<li><span><code>s</code> and <code>t</code> only contain lowercase letters and <code>&#39;#&#39;</code> characters.</span></li>\n</ul>\n\n<p>&nbsp;</p>\n<p><strong>Follow up:</strong> Can you solve it in <code>O(n)</code> time and <code>O(1)</code> space?</p>\n",
            "translatedTitle": "比较含退格的字符串",
            "translatedContent": "<p>给定 <code>s</code> 和 <code>t</code> 两个字符串，当它们分别被输入到空白的文本编辑器后，如果两者相等，返回 <code>true</code> 。<code>#</code> 代表退格字符。</p>\n\n<p><strong>注意：</strong>如果对空文本输入退格字符，文本继续为空。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>s = \"ab#c\", t = \"ad#c\"\n<strong>输出：</strong>true\n<strong>解释：</strong>s 和 t 都会变成 \"ac\"。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>s = \"ab##\", t = \"c#d#\"\n<strong>输出：</strong>true\n<strong>解释：</strong>s 和 t 都会变成 \"\"。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>s = \"a#c\", t = \"b\"\n<strong>输出：</strong>false\n<strong>解释：</strong>s 会变成 \"c\"，但 t 仍然是 \"b\"。</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= s.length, t.length &lt;= 200</code></li>\n\t<li><code>s</code> 和 <code>t</code> 只含有小写字母以及字符 <code>'#'</code></li>\n</ul>\n\n<p>&nbsp;</p>\n\n<p><strong>进阶：</strong></p>\n\n<ul>\n\t<li>你可以用 <code>O(n)</code> 的时间复杂度和 <code>O(1)</code> 的空间复杂度解决该问题吗？</li>\n</ul>\n",
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