leetcode-problemset/leetcode-cn/originData/last-visited-integers.json

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        "question": {
            "questionId": "3164",
            "questionFrontendId": "2899",
            "categoryTitle": "Algorithms",
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            "title": "Last Visited Integers",
            "titleSlug": "last-visited-integers",
            "content": "<p>Given a <strong>0-indexed</strong> array of strings <code>words</code> where <code>words[i]</code> is either a positive integer represented as a string or the string <code>&quot;prev&quot;</code>.</p>\n\n<p>Start iterating from the beginning of the array; for every <code>&quot;prev&quot;</code> string seen in <code>words</code>, find the <strong>last visited integer</strong> in <code>words</code> which is defined as follows:</p>\n\n<ul>\n\t<li>Let <code>k</code> be the number of consecutive <code>&quot;prev&quot;</code> strings seen so far (containing the current string). Let <code>nums</code> be the <strong>0-indexed </strong>array of <strong>integers</strong> seen so far and <code>nums_reverse</code> be the reverse of <code>nums</code>, then the integer at <code>(k - 1)<sup>th</sup></code> index of <code>nums_reverse</code> will be the <strong>last visited integer</strong> for this <code>&quot;prev&quot;</code>.</li>\n\t<li>If <code>k</code> is <strong>greater</strong> than the total visited integers, then the last visited integer will be <code>-1</code>.</li>\n</ul>\n\n<p>Return <em>an integer array containing the last visited integers.</em></p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> words = [&quot;1&quot;,&quot;2&quot;,&quot;prev&quot;,&quot;prev&quot;,&quot;prev&quot;]\n<strong>Output:</strong> [2,1,-1]\n<strong>Explanation:</strong> \nFor &quot;prev&quot; at index = 2, last visited integer will be 2 as here the number of consecutive &quot;prev&quot; strings is 1, and in the array reverse_nums, 2 will be the first element.\nFor &quot;prev&quot; at index = 3, last visited integer will be 1 as there are a total of two consecutive &quot;prev&quot; strings including this &quot;prev&quot; which are visited, and 1 is the second last visited integer.\nFor &quot;prev&quot; at index = 4, last visited integer will be -1 as there are a total of three consecutive &quot;prev&quot; strings including this &quot;prev&quot; which are visited, but the total number of integers visited is two.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> words = [&quot;1&quot;,&quot;prev&quot;,&quot;2&quot;,&quot;prev&quot;,&quot;prev&quot;]\n<strong>Output:</strong> [1,2,1]\n<strong>Explanation:</strong>\nFor &quot;prev&quot; at index = 1, last visited integer will be 1.\nFor &quot;prev&quot; at index = 3, last visited integer will be 2.\nFor &quot;prev&quot; at index = 4, last visited integer will be 1 as there are a total of two consecutive &quot;prev&quot; strings including this &quot;prev&quot; which are visited, and 1 is the second last visited integer.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= words.length &lt;= 100</code></li>\n\t<li><code>words[i] == &quot;prev&quot;</code> or <code>1 &lt;= int(words[i]) &lt;= 100</code></li>\n</ul>\n",
            "translatedTitle": "上一个遍历的整数",
            "translatedContent": "<p>给你一个下标从 <strong>0</strong>&nbsp;开始的字符串数组&nbsp;<code>words</code>&nbsp;，其中&nbsp;<code>words[i]</code>&nbsp;要么是一个字符串形式的正整数，要么是字符串&nbsp;<code>\"prev\"</code>&nbsp;。</p>\n\n<p>我们从数组的开头开始遍历，对于 <code>words</code>&nbsp;中的每个&nbsp;<code>\"prev\"</code>&nbsp;字符串，找到 <code>words</code>&nbsp;中的 <strong>上一个遍历的整数</strong>&nbsp;，定义如下：</p>\n\n<ul>\n\t<li><code>k</code>&nbsp;表示到当前位置为止的连续&nbsp;<code>\"prev\"</code>&nbsp;字符串数目（包含当前字符串），令下标从&nbsp;<strong>0</strong>&nbsp;开始的&nbsp;<strong>整数</strong> 数组&nbsp;<code>nums</code>&nbsp;表示目前为止遍历过的所有整数，同时用&nbsp;<code>nums_reverse</code>&nbsp;表示&nbsp;<code>nums</code>&nbsp;反转得到的数组，那么当前 <code>\"prev\"</code>&nbsp;对应的 <strong>上一个遍历的整数</strong>&nbsp;是&nbsp;<code>nums_reverse</code>&nbsp;数组中下标为 <code>(k - 1)</code>&nbsp;的整数。</li>\n\t<li>如果&nbsp;<code>k</code>&nbsp;比目前为止遍历过的整数数目 <strong>更多</strong>&nbsp;，那么上一个遍历的整数为&nbsp;<code>-1</code>&nbsp;。</li>\n</ul>\n\n<p>请你返回一个整数数组，包含所有上一个遍历的整数。</p>\n\n<p>&nbsp;</p>\n\n<p><strong class=\"example\">示例 1：</strong></p>\n\n<pre>\n<b>输入：</b><code>words</code> = [\"1\",\"2\",\"prev\",\"prev\",\"prev\"]\n<b>输出：</b>[2,1,-1]\n<b>解释：</b>\n对于下标为 2 处的 \"prev\" ，上一个遍历的整数是 2 ，因为连续 \"prev\" 数目为 1 ，同时在数组 reverse_nums 中，第一个元素是 2 。\n对于下标为 3 处的 \"prev\" ，上一个遍历的整数是 1 ，因为连续 \"prev\" 数目为 2 ，同时在数组 reverse_nums 中，第二个元素是 1 。\n对于下标为 4 处的 \"prev\" ，上一个遍历的整数是 -1 ，因为连续 \"prev\" 数目为 3 ，但总共只遍历过 2 个整数。\n</pre>\n\n<p><strong class=\"example\">示例 2：</strong></p>\n\n<pre>\n<b>输入：</b><code>words</code> = [\"1\",\"prev\",\"2\",\"prev\",\"prev\"]\n<b>输出：</b>[1,2,1]\n<strong>解释：</strong>\n对于下标为 1 处的 \"prev\" ，上一个遍历的整数是 1 。\n对于下标为 3 处的 \"prev\" ，上一个遍历的整数是 2 。\n对于下标为 4 处的 \"prev\" ，上一个遍历的整数是 1 ，因为连续 \"prev\"<strong>&nbsp;</strong>数目为 2 ，同时在数组 reverse_nums 中，第二个元素是 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;= words.length &lt;= 100</code></li>\n\t<li><code>words[i] == \"prev\"</code>&nbsp;或&nbsp;<code>1 &lt;= int(words[i]) &lt;= 100</code></li>\n</ul>\n",
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                    "code": "class Solution {\npublic:\n    vector<int> lastVisitedIntegers(vector<string>& words) {\n        \n    }\n};",
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                    "code": "class Solution(object):\n    def lastVisitedIntegers(self, words):\n        \"\"\"\n        :type words: List[str]\n        :rtype: List[int]\n        \"\"\"",
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                    "code": "/**\n * Note: The returned array must be malloced, assume caller calls free().\n */\nint* lastVisitedIntegers(char** words, int wordsSize, int* returnSize) {\n    \n}",
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                    "code": "public class Solution {\n    public IList<int> LastVisitedIntegers(IList<string> words) {\n\n    }\n}",
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                    "code": "class Solution {\n\n    /**\n     * @param String[] $words\n     * @return Integer[]\n     */\n    function lastVisitedIntegers($words) {\n\n    }\n}",
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                    "code": "class Solution {\n    func lastVisitedIntegers(_ words: [String]) -> [Int] {\n\n    }\n}",
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                    "code": "class Solution {\n    fun lastVisitedIntegers(words: List<String>): List<Int> {\n\n    }\n}",
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                    "code": "class Solution {\n  List<int> lastVisitedIntegers(List<String> words) {\n    \n  }\n}",
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                    "code": "func lastVisitedIntegers(words []string) []int {\n\n}",
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                    "code": "# @param {String[]} words\n# @return {Integer[]}\ndef last_visited_integers(words)\n\nend",
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