leetcode-problemset/leetcode-cn/originData/shifting-letters.json

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            "categoryTitle": "Algorithms",
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            "title": "Shifting Letters",
            "titleSlug": "shifting-letters",
            "content": "<p>You are given a string <code>s</code> of lowercase English letters and an integer array <code>shifts</code> of the same length.</p>\n\n<p>Call the <code>shift()</code> of a letter, the next letter in the alphabet, (wrapping around so that <code>&#39;z&#39;</code> becomes <code>&#39;a&#39;</code>).</p>\n\n<ul>\n\t<li>For example, <code>shift(&#39;a&#39;) = &#39;b&#39;</code>, <code>shift(&#39;t&#39;) = &#39;u&#39;</code>, and <code>shift(&#39;z&#39;) = &#39;a&#39;</code>.</li>\n</ul>\n\n<p>Now for each <code>shifts[i] = x</code>, we want to shift the first <code>i + 1</code> letters of <code>s</code>, <code>x</code> times.</p>\n\n<p>Return <em>the final string after all such shifts to s are applied</em>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;abc&quot;, shifts = [3,5,9]\n<strong>Output:</strong> &quot;rpl&quot;\n<strong>Explanation:</strong> We start with &quot;abc&quot;.\nAfter shifting the first 1 letters of s by 3, we have &quot;dbc&quot;.\nAfter shifting the first 2 letters of s by 5, we have &quot;igc&quot;.\nAfter shifting the first 3 letters of s by 9, we have &quot;rpl&quot;, the answer.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;aaa&quot;, shifts = [1,2,3]\n<strong>Output:</strong> &quot;gfd&quot;\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>s</code> consists of lowercase English letters.</li>\n\t<li><code>shifts.length == s.length</code></li>\n\t<li><code>0 &lt;= shifts[i] &lt;= 10<sup>9</sup></code></li>\n</ul>\n",
            "translatedTitle": "字母移位",
            "translatedContent": "<p>有一个由小写字母组成的字符串 <code>s</code>，和一个长度相同的整数数组 <code>shifts</code>。</p>\n\n<p>我们将字母表中的下一个字母称为原字母的 <em>移位</em>&nbsp;<code>shift()</code>&nbsp;（由于字母表是环绕的， <code>'z'</code>&nbsp;将会变成&nbsp;<code>'a'</code>）。</p>\n\n<ul>\n\t<li>例如，<code>shift('a') = 'b'<font color=\"#333333\"><font face=\"Helvetica Neue, Helvetica, Arial, sans-serif\"><span style=\"font-size:14px\"><span style=\"background-color:#ffffff\">,&nbsp;</span></span></font></font></code><code>shift('t') = 'u'</code>,&nbsp;以及&nbsp;<code>shift('z') = 'a'</code>。</li>\n</ul>\n\n<p>对于每个&nbsp;<code>shifts[i] = x</code>&nbsp;， 我们会将 <code>s</code>&nbsp;中的前&nbsp;<code>i + 1</code>&nbsp;个字母移位&nbsp;<code>x</code>&nbsp;次。</p>\n\n<p>返回 <em>将所有这些移位都应用到 <code>s</code> 后最终得到的字符串</em> 。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>s = \"abc\", shifts = [3,5,9]\n<strong>输出：</strong>\"rpl\"\n<strong>解释： </strong>\n我们以 \"abc\" 开始。\n将 S 中的第 1 个字母移位 3 次后，我们得到 \"dbc\"。\n再将 S 中的前 2 个字母移位 5 次后，我们得到 \"igc\"。\n最后将 S 中的这 3 个字母移位 9 次后，我们得到答案 \"rpl\"。\n</pre>\n\n<p><strong>示例 2:</strong></p>\n\n<pre>\n<strong>输入:</strong> s = \"aaa\", shifts = [1,2,3]\n<strong>输出:</strong> \"gfd\"\n</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 &lt;= 10<sup>5</sup></code></li>\n\t<li><code>s</code>&nbsp;由小写英文字母组成</li>\n\t<li><code>shifts.length == s.length</code></li>\n\t<li><code>0 &lt;= shifts[i] &lt;= 10<sup>9</sup></code></li>\n</ul>\n<span style=\"display:block\"><span style=\"height:0px\"><span style=\"position:absolute\">​​​​​​</span></span></span>",
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