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# 力扣题库（完整版）
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> 最后更新日期： **2022.10.26**
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> 最后更新日期： **2022.11.01**
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>
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> 使用脚本前请务必仔细完整阅读本 `README.md` 文件
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leetcode-cn/originData/next-greater-element-iv.json
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leetcode-cn/originData/next-greater-element-iv.json
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leetcode-cn/originData/words-within-two-edits-of-dictionary.json
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leetcode-cn/originData/words-within-two-edits-of-dictionary.json
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<p>给你一个下标从 <strong>0</strong> 开始的非负整数数组 <code>nums</code> 。对于 <code>nums</code> 中每一个整数，你必须找到对应元素的 <strong>第二大</strong> 整数。</p>
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<p>如果 <code>nums[j]</code> 满足以下条件，那么我们称它为 <code>nums[i]</code> 的 <strong>第二大</strong> 整数：</p>
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<ul>
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<li><code>j > i</code></li>
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<li><code>nums[j] > nums[i]</code></li>
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<li>恰好存在 <strong>一个</strong> <code>k</code> 满足 <code>i < k < j</code> 且 <code>nums[k] > nums[i]</code> 。</li>
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</ul>
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<p>如果不存在 <code>nums[j]</code> ，那么第二大整数为 <code>-1</code> 。</p>
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<ul>
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<li>比方说，数组 <code>[1, 2, 4, 3]</code> 中，<code>1</code> 的第二大整数是 <code>4</code> ，<code>2</code> 的第二大整数是 <code>3</code> ，<code>3</code> 和 <code>4</code> 的第二大整数是 <code>-1</code> 。</li>
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</ul>
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<p>请你返回一个整数数组<em> </em><code>answer</code> ，其中<em> </em><code>answer[i]</code>是<em> </em><code>nums[i]</code> 的第二大整数。</p>
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<p> </p>
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<p><strong>示例 1：</strong></p>
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<pre>
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<b>输入：</b>nums = [2,4,0,9,6]
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<b>输出：</b>[9,6,6,-1,-1]
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<strong>解释：</strong>
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下标为 0 处：2 的右边，4 是大于 2 的第一个整数，9 是第二个大于 2 的整数。
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下标为 1 处：4 的右边，9 是大于 4 的第一个整数，6 是第二个大于 4 的整数。
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下标为 2 处：0 的右边，9 是大于 0 的第一个整数，6 是第二个大于 0 的整数。
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下标为 3 处：右边不存在大于 9 的整数，所以第二大整数为 -1 。
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下标为 4 处：右边不存在大于 6 的整数，所以第二大整数为 -1 。
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所以我们返回 [9,6,6,-1,-1] 。
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</pre>
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<p><strong>示例 2：</strong></p>
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<pre>
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<strong>输入：</strong>nums = [3,3]
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<b>输出：</b>[-1,-1]
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<strong>解释：</strong>
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由于每个数右边都没有更大的数，所以我们返回 [-1,-1] 。
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</pre>
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<p> </p>
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<p><strong>提示：</strong></p>
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<ul>
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<li><code>1 <= nums.length <= 10<sup>5</sup></code></li>
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<li><code>0 <= nums[i] <= 10<sup>9</sup></code></li>
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</ul>
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<p>给你两个字符串数组 <code>queries</code> 和 <code>dictionary</code> 。数组中所有单词都只包含小写英文字母，且长度都相同。</p>
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<p>一次 <strong>编辑</strong> 中，你可以从 <code>queries</code> 中选择一个单词，将任意一个字母修改成任何其他字母。从 <code>queries</code> 中找到所有满足以下条件的字符串：<strong>不超过</strong> 两次编辑内，字符串与 <code>dictionary</code> 中某个字符串相同。</p>
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<p>请你返回<em> </em><code>queries</code> 中的单词列表，这些单词距离 <code>dictionary</code> 中的单词 <strong>编辑次数</strong> 不超过 <strong>两次</strong> 。单词返回的顺序需要与 <code>queries</code> 中原本顺序相同。</p>
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<p> </p>
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<p><strong>示例 1：</strong></p>
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<pre><b>输入：</b>queries = ["word","note","ants","wood"], dictionary = ["wood","joke","moat"]
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<b>输出：</b>["word","note","wood"]
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<strong>解释：</strong>
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- 将 "word" 中的 'r' 换成 'o' ，得到 dictionary 中的单词 "wood" 。
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- 将 "note" 中的 'n' 换成 'j' 且将 't' 换成 'k' ，得到 "joke" 。
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- "ants" 需要超过 2 次编辑才能得到 dictionary 中的单词。
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- "wood" 不需要修改（0 次编辑），就得到 dictionary 中相同的单词。
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所以我们返回 ["word","note","wood"] 。
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</pre>
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<p><strong>示例 2：</strong></p>
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<pre><b>输入：</b>queries = ["yes"], dictionary = ["not"]
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<b>输出：</b>[]
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<strong>解释：</strong>
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"yes" 需要超过 2 次编辑才能得到 "not" 。
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所以我们返回空数组。
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</pre>
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<p> </p>
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<p><strong>提示：</strong></p>
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<ul>
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<li><code>1 <= queries.length, dictionary.length <= 100</code></li>
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<li><code>n == queries[i].length == dictionary[j].length</code></li>
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<li><code>1 <= n <= 100</code></li>
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<li>所有 <code>queries[i]</code> 和 <code>dictionary[j]</code> 都只包含小写英文字母。</li>
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</ul>
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<p>You are given a <strong>0-indexed</strong> array of non-negative integers <code>nums</code>. For each integer in <code>nums</code>, you must find its respective <strong>second greater</strong> integer.</p>
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<p>The <strong>second greater</strong> integer of <code>nums[i]</code> is <code>nums[j]</code> such that:</p>
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<ul>
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<li><code>j > i</code></li>
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<li><code>nums[j] > nums[i]</code></li>
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<li>There exists <strong>exactly one</strong> index <code>k</code> such that <code>nums[k] > nums[i]</code> and <code>i < k < j</code>.</li>
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</ul>
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<p>If there is no such <code>nums[j]</code>, the second greater integer is considered to be <code>-1</code>.</p>
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<ul>
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<li>For example, in the array <code>[1, 2, 4, 3]</code>, the second greater integer of <code>1</code> is <code>4</code>, <code>2</code> is <code>3</code>, and that of <code>3</code> and <code>4</code> is <code>-1</code>.</li>
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</ul>
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<p>Return<em> an integer array </em><code>answer</code><em>, where </em><code>answer[i]</code><em> is the second greater integer of </em><code>nums[i]</code><em>.</em></p>
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<p> </p>
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<p><strong class="example">Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> nums = [2,4,0,9,6]
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<strong>Output:</strong> [9,6,6,-1,-1]
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<strong>Explanation:</strong>
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0th index: 4 is the first integer greater than 2, and 9 is the second integer greater than 2, to the right of 2.
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1st index: 9 is the first, and 6 is the second integer greater than 4, to the right of 4.
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2nd index: 9 is the first, and 6 is the second integer greater than 0, to the right of 0.
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3rd index: There is no integer greater than 9 to its right, so the second greater integer is considered to be -1.
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4th index: There is no integer greater than 6 to its right, so the second greater integer is considered to be -1.
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Thus, we return [9,6,6,-1,-1].
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</pre>
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<p><strong class="example">Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> nums = [3,3]
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<strong>Output:</strong> [-1,-1]
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<strong>Explanation:</strong>
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We return [-1,-1] since neither integer has any integer greater than it.
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</pre>
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<p> </p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 <= nums.length <= 10<sup>5</sup></code></li>
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<li><code>0 <= nums[i] <= 10<sup>9</sup></code></li>
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</ul>
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<p>You are given two string arrays, <code>queries</code> and <code>dictionary</code>. All words in each array comprise of lowercase English letters and have the same length.</p>
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<p>In one <strong>edit</strong> you can take a word from <code>queries</code>, and change any letter in it to any other letter. Find all words from <code>queries</code> that, after a <strong>maximum</strong> of two edits, equal some word from <code>dictionary</code>.</p>
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<p>Return<em> a list of all words from </em><code>queries</code><em>, </em><em>that match with some word from </em><code>dictionary</code><em> after a maximum of <strong>two edits</strong></em>. Return the words in the <strong>same order</strong> they appear in <code>queries</code>.</p>
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<p> </p>
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<p><strong class="example">Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> queries = ["word","note","ants","wood"], dictionary = ["wood","joke","moat"]
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<strong>Output:</strong> ["word","note","wood"]
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<strong>Explanation:</strong>
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- Changing the 'r' in "word" to 'o' allows it to equal the dictionary word "wood".
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- Changing the 'n' to 'j' and the 't' to 'k' in "note" changes it to "joke".
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- It would take more than 2 edits for "ants" to equal a dictionary word.
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- "wood" can remain unchanged (0 edits) and match the corresponding dictionary word.
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Thus, we return ["word","note","wood"].
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</pre>
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<p><strong class="example">Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> queries = ["yes"], dictionary = ["not"]
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<strong>Output:</strong> []
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<strong>Explanation:</strong>
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Applying any two edits to "yes" cannot make it equal to "not". Thus, we return an empty array.
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</pre>
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<p> </p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 <= queries.length, dictionary.length <= 100</code></li>
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<li><code>n == queries[i].length == dictionary[j].length</code></li>
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<li><code>1 <= n <= 100</code></li>
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<li>All <code>queries[i]</code> and <code>dictionary[j]</code> are composed of lowercase English letters.</li>
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</ul>
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