leetcode-problemset/leetcode-cn/originData/freedom-trail.json

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            "title": "Freedom Trail",
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            "content": "<p>In the video game Fallout 4, the quest <strong>&quot;Road to Freedom&quot;</strong> requires players to reach a metal dial called the <strong>&quot;Freedom Trail Ring&quot;</strong> and use the dial to spell a specific keyword to open the door.</p>\n\n<p>Given a string <code>ring</code> that represents the code engraved on the outer ring and another string <code>key</code> that represents the keyword that needs to be spelled, return <em>the minimum number of steps to spell all the characters in the keyword</em>.</p>\n\n<p>Initially, the first character of the ring is aligned at the <code>&quot;12:00&quot;</code> direction. You should spell all the characters in <code>key</code> one by one by rotating <code>ring</code> clockwise or anticlockwise to make each character of the string key aligned at the <code>&quot;12:00&quot;</code> direction and then by pressing the center button.</p>\n\n<p>At the stage of rotating the ring to spell the key character <code>key[i]</code>:</p>\n\n<ol>\n\t<li>You can rotate the ring clockwise or anticlockwise by one place, which counts as <strong>one step</strong>. The final purpose of the rotation is to align one of <code>ring</code>&#39;s characters at the <code>&quot;12:00&quot;</code> direction, where this character must equal <code>key[i]</code>.</li>\n\t<li>If the character <code>key[i]</code> has been aligned at the <code>&quot;12:00&quot;</code> direction, press the center button to spell, which also counts as <strong>one step</strong>. After the pressing, you could begin to spell the next character in the key (next stage). Otherwise, you have finished all the spelling.</li>\n</ol>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n<img src=\"https://assets.leetcode.com/uploads/2018/10/22/ring.jpg\" style=\"width: 450px; height: 450px;\" />\n<pre>\n<strong>Input:</strong> ring = &quot;godding&quot;, key = &quot;gd&quot;\n<strong>Output:</strong> 4\n<strong>Explanation:</strong>\nFor the first key character &#39;g&#39;, since it is already in place, we just need 1 step to spell this character. \nFor the second key character &#39;d&#39;, we need to rotate the ring &quot;godding&quot; anticlockwise by two steps to make it become &quot;ddinggo&quot;.\nAlso, we need 1 more step for spelling.\nSo the final output is 4.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> ring = &quot;godding&quot;, key = &quot;godding&quot;\n<strong>Output:</strong> 13\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= ring.length, key.length &lt;= 100</code></li>\n\t<li><code>ring</code> and <code>key</code> consist of only lower case English letters.</li>\n\t<li>It is guaranteed that <code>key</code> could always be spelled by rotating <code>ring</code>.</li>\n</ul>\n",
            "translatedTitle": "自由之路",
            "translatedContent": "<p>电子游戏“辐射4”中，任务 <strong>“通向自由”</strong> 要求玩家到达名为 “<strong>Freedom Trail Ring”</strong> 的金属表盘，并使用表盘拼写特定关键词才能开门。</p>\n\n<p>给定一个字符串&nbsp;<code>ring</code>&nbsp;，表示刻在外环上的编码；给定另一个字符串&nbsp;<code>key</code>&nbsp;，表示需要拼写的关键词。您需要算出能够拼写关键词中所有字符的<strong>最少</strong>步数。</p>\n\n<p>最初，<strong>ring&nbsp;</strong>的第一个字符与 <code>12:00</code> 方向对齐。您需要顺时针或逆时针旋转 <code>ring</code> 以使&nbsp;<strong>key&nbsp;</strong>的一个字符在 <code>12:00</code> 方向对齐，然后按下中心按钮，以此逐个拼写完&nbsp;<strong><code>key</code>&nbsp;</strong>中的所有字符。</p>\n\n<p>旋转&nbsp;<code>ring</code><strong>&nbsp;</strong>拼出 key 字符&nbsp;<code>key[i]</code><strong>&nbsp;</strong>的阶段中：</p>\n\n<ol>\n\t<li>您可以将&nbsp;<strong>ring&nbsp;</strong>顺时针或逆时针旋转&nbsp;<strong>一个位置&nbsp;</strong>，计为1步。旋转的最终目的是将字符串&nbsp;<strong><code>ring</code>&nbsp;</strong>的一个字符与 <code>12:00</code> 方向对齐，并且这个字符必须等于字符&nbsp;<strong><code>key[i]</code> 。</strong></li>\n\t<li>如果字符&nbsp;<strong><code>key[i]</code>&nbsp;</strong>已经对齐到12:00方向，您需要按下中心按钮进行拼写，这也将算作&nbsp;<strong>1 步</strong>。按完之后，您可以开始拼写&nbsp;<strong>key&nbsp;</strong>的下一个字符（下一阶段）, 直至完成所有拼写。</li>\n</ol>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<p><img src=\"https://assets.leetcode.com/uploads/2018/10/22/ring.jpg\" style=\"height: 450px; width: 450px;\" /></p>\n\n<center>&nbsp;</center>\n\n<pre>\n<strong>输入:</strong> ring = \"godding\", key = \"gd\"\n<strong>输出:</strong> 4\n<strong>解释:</strong>\n 对于 key 的第一个字符 'g'，已经在正确的位置, 我们只需要1步来拼写这个字符。 \n 对于 key 的第二个字符 'd'，我们需要逆时针旋转 ring \"godding\" 2步使它变成 \"ddinggo\"。\n 当然, 我们还需要1步进行拼写。\n 因此最终的输出是 4。\n</pre>\n\n<p><strong>示例 2:</strong></p>\n\n<pre>\n<strong>输入:</strong> ring = \"godding\", key = \"godding\"\n<strong>输出:</strong> 13\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= ring.length, key.length &lt;= 100</code></li>\n\t<li><code>ring</code>&nbsp;和&nbsp;<code>key</code>&nbsp;只包含小写英文字母</li>\n\t<li><strong>保证</strong> 字符串&nbsp;<code>key</code>&nbsp;一定可以由字符串 &nbsp;<code>ring</code>&nbsp;旋转拼出</li>\n</ul>\n",
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