leetcode-problemset/leetcode-cn/originData/next-greater-element-i.json

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            "title": "Next Greater Element I",
            "titleSlug": "next-greater-element-i",
            "content": "<p>The <strong>next greater element</strong> of some element <code>x</code> in an array is the <strong>first greater</strong> element that is <strong>to the right</strong> of <code>x</code> in the same array.</p>\n\n<p>You are given two <strong>distinct 0-indexed</strong> integer arrays <code>nums1</code> and <code>nums2</code>, where <code>nums1</code> is a subset of <code>nums2</code>.</p>\n\n<p>For each <code>0 &lt;= i &lt; nums1.length</code>, find the index <code>j</code> such that <code>nums1[i] == nums2[j]</code> and determine the <strong>next greater element</strong> of <code>nums2[j]</code> in <code>nums2</code>. If there is no next greater element, then the answer for this query is <code>-1</code>.</p>\n\n<p>Return <em>an array </em><code>ans</code><em> of length </em><code>nums1.length</code><em> such that </em><code>ans[i]</code><em> is the <strong>next greater element</strong> as described above.</em></p>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums1 = [4,1,2], nums2 = [1,3,4,2]\n<strong>Output:</strong> [-1,3,-1]\n<strong>Explanation:</strong> The next greater element for each value of nums1 is as follows:\n- 4 is underlined in nums2 = [1,3,<u>4</u>,2]. There is no next greater element, so the answer is -1.\n- 1 is underlined in nums2 = [<u>1</u>,3,4,2]. The next greater element is 3.\n- 2 is underlined in nums2 = [1,3,4,<u>2</u>]. There is no next greater element, so the answer is -1.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums1 = [2,4], nums2 = [1,2,3,4]\n<strong>Output:</strong> [3,-1]\n<strong>Explanation:</strong> The next greater element for each value of nums1 is as follows:\n- 2 is underlined in nums2 = [1,<u>2</u>,3,4]. The next greater element is 3.\n- 4 is underlined in nums2 = [1,2,3,<u>4</u>]. There is no next greater element, so the answer is -1.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= nums1.length &lt;= nums2.length &lt;= 1000</code></li>\n\t<li><code>0 &lt;= nums1[i], nums2[i] &lt;= 10<sup>4</sup></code></li>\n\t<li>All integers in <code>nums1</code> and <code>nums2</code> are <strong>unique</strong>.</li>\n\t<li>All the integers of <code>nums1</code> also appear in <code>nums2</code>.</li>\n</ul>\n\n<p>&nbsp;</p>\n<strong>Follow up:</strong> Could you find an <code>O(nums1.length + nums2.length)</code> solution?",
            "translatedTitle": "下一个更大元素 I",
            "translatedContent": "<p><code>nums1</code>&nbsp;中数字&nbsp;<code>x</code>&nbsp;的 <strong>下一个更大元素</strong> 是指&nbsp;<code>x</code>&nbsp;在&nbsp;<code>nums2</code> 中对应位置 <strong>右侧</strong> 的 <strong>第一个</strong> 比&nbsp;<code>x</code><strong>&nbsp;</strong>大的元素。</p>\n\n<p>给你两个<strong> 没有重复元素</strong> 的数组&nbsp;<code>nums1</code> 和&nbsp;<code>nums2</code> ，下标从 <strong>0</strong> 开始计数，其中<code>nums1</code>&nbsp;是&nbsp;<code>nums2</code>&nbsp;的子集。</p>\n\n<p>对于每个 <code>0 &lt;= i &lt; nums1.length</code> ，找出满足 <code>nums1[i] == nums2[j]</code> 的下标 <code>j</code> ，并且在 <code>nums2</code> 确定 <code>nums2[j]</code> 的 <strong>下一个更大元素</strong> 。如果不存在下一个更大元素，那么本次查询的答案是 <code>-1</code> 。</p>\n\n<p>返回一个长度为&nbsp;<code>nums1.length</code> 的数组<em> </em><code>ans</code><em> </em>作为答案，满足<em> </em><code>ans[i]</code><em> </em>是如上所述的 <strong>下一个更大元素</strong> 。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums1 = [4,1,2], nums2 = [1,3,4,2].\n<strong>输出：</strong>[-1,3,-1]\n<strong>解释：</strong>nums1 中每个值的下一个更大元素如下所述：\n- 4 ，用加粗斜体标识，nums2 = [1,3,<strong>4</strong>,2]。不存在下一个更大元素，所以答案是 -1 。\n- 1 ，用加粗斜体标识，nums2 = [<em><strong>1</strong></em>,3,4,2]。下一个更大元素是 3 。\n- 2 ，用加粗斜体标识，nums2 = [1,3,4,<em><strong>2</strong></em>]。不存在下一个更大元素，所以答案是 -1 。</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums1 = [2,4], nums2 = [1,2,3,4].\n<strong>输出：</strong>[3,-1]\n<strong>解释：</strong>nums1 中每个值的下一个更大元素如下所述：\n- 2 ，用加粗斜体标识，nums2 = [1,<em><strong>2</strong></em>,3,4]。下一个更大元素是 3 。\n- 4 ，用加粗斜体标识，nums2 = [1,2,3,<em><strong>4</strong></em>]。不存在下一个更大元素，所以答案是 -1 。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= nums1.length &lt;= nums2.length &lt;= 1000</code></li>\n\t<li><code>0 &lt;= nums1[i], nums2[i] &lt;= 10<sup>4</sup></code></li>\n\t<li><code>nums1</code>和<code>nums2</code>中所有整数 <strong>互不相同</strong></li>\n\t<li><code>nums1</code> 中的所有整数同样出现在 <code>nums2</code> 中</li>\n</ul>\n\n<p>&nbsp;</p>\n\n<p><strong>进阶：</strong>你可以设计一个时间复杂度为 <code>O(nums1.length + nums2.length)</code> 的解决方案吗？</p>\n",
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