leetcode-problemset/leetcode-cn/originData/minimum-cost-homecoming-of-a-robot-in-a-grid.json

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            "title": "Minimum Cost Homecoming of a Robot in a Grid",
            "titleSlug": "minimum-cost-homecoming-of-a-robot-in-a-grid",
            "content": "<p>There is an <code>m x n</code> grid, where <code>(0, 0)</code> is the top-left cell and <code>(m - 1, n - 1)</code> is the bottom-right cell. You are given an integer array <code>startPos</code> where <code>startPos = [start<sub>row</sub>, start<sub>col</sub>]</code> indicates that <strong>initially</strong>, a <strong>robot</strong> is at the cell <code>(start<sub>row</sub>, start<sub>col</sub>)</code>. You are also given an integer array <code>homePos</code> where <code>homePos = [home<sub>row</sub>, home<sub>col</sub>]</code> indicates that its <strong>home</strong> is at the cell <code>(home<sub>row</sub>, home<sub>col</sub>)</code>.</p>\n\n<p>The robot needs to go to its home. It can move one cell in four directions: <strong>left</strong>, <strong>right</strong>, <strong>up</strong>, or <strong>down</strong>, and it can not move outside the boundary. Every move incurs some cost. You are further given two <strong>0-indexed</strong> integer arrays: <code>rowCosts</code> of length <code>m</code> and <code>colCosts</code> of length <code>n</code>.</p>\n\n<ul>\n\t<li>If the robot moves <strong>up</strong> or <strong>down</strong> into a cell whose <strong>row</strong> is <code>r</code>, then this move costs <code>rowCosts[r]</code>.</li>\n\t<li>If the robot moves <strong>left</strong> or <strong>right</strong> into a cell whose <strong>column</strong> is <code>c</code>, then this move costs <code>colCosts[c]</code>.</li>\n</ul>\n\n<p>Return <em>the <strong>minimum total cost</strong> for this robot to return home</em>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/10/11/eg-1.png\" style=\"width: 282px; height: 217px;\" />\n<pre>\n<strong>Input:</strong> startPos = [1, 0], homePos = [2, 3], rowCosts = [5, 4, 3], colCosts = [8, 2, 6, 7]\n<strong>Output:</strong> 18\n<strong>Explanation:</strong> One optimal path is that:\nStarting from (1, 0)\n-&gt; It goes down to (<u><strong>2</strong></u>, 0). This move costs rowCosts[2] = 3.\n-&gt; It goes right to (2, <u><strong>1</strong></u>). This move costs colCosts[1] = 2.\n-&gt; It goes right to (2, <u><strong>2</strong></u>). This move costs colCosts[2] = 6.\n-&gt; It goes right to (2, <u><strong>3</strong></u>). This move costs colCosts[3] = 7.\nThe total cost is 3 + 2 + 6 + 7 = 18</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> startPos = [0, 0], homePos = [0, 0], rowCosts = [5], colCosts = [26]\n<strong>Output:</strong> 0\n<strong>Explanation:</strong> The robot is already at its home. Since no moves occur, the total cost is 0.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>m == rowCosts.length</code></li>\n\t<li><code>n == colCosts.length</code></li>\n\t<li><code>1 &lt;= m, n &lt;= 10<sup>5</sup></code></li>\n\t<li><code>0 &lt;= rowCosts[r], colCosts[c] &lt;= 10<sup>4</sup></code></li>\n\t<li><code>startPos.length == 2</code></li>\n\t<li><code>homePos.length == 2</code></li>\n\t<li><code>0 &lt;= start<sub>row</sub>, home<sub>row</sub> &lt; m</code></li>\n\t<li><code>0 &lt;= start<sub>col</sub>, home<sub>col</sub> &lt; n</code></li>\n</ul>\n",
            "translatedTitle": "网格图中机器人回家的最小代价",
            "translatedContent": "<p>给你一个&nbsp;<code>m x n</code>&nbsp;的网格图，其中&nbsp;<code>(0, 0)</code>&nbsp;是最左上角的格子，<code>(m - 1, n - 1)</code>&nbsp;是最右下角的格子。给你一个整数数组&nbsp;<code>startPos</code>&nbsp;，<code>startPos = [start<sub>row</sub>, start<sub>col</sub>]</code>&nbsp;表示 <strong>初始</strong>&nbsp;有一个 <strong>机器人</strong>&nbsp;在格子&nbsp;<code>(start<sub>row</sub>, start<sub>col</sub>)</code>&nbsp;处。同时给你一个整数数组&nbsp;<code>homePos</code>&nbsp;，<code>homePos = [home<sub>row</sub>, home<sub>col</sub>]</code>&nbsp;表示机器人的 <strong>家</strong>&nbsp;在格子&nbsp;<code>(home<sub>row</sub>, home<sub>col</sub>)</code>&nbsp;处。</p>\n\n<p>机器人需要回家。每一步它可以往四个方向移动：<strong>上</strong>，<strong>下</strong>，<strong>左</strong>，<strong>右</strong>，同时机器人不能移出边界。每一步移动都有一定代价。再给你两个下标从&nbsp;<strong>0</strong>&nbsp;开始的额整数数组：长度为&nbsp;<code>m</code>&nbsp;的数组&nbsp;<code>rowCosts</code> &nbsp;和长度为 <code>n</code>&nbsp;的数组&nbsp;<code>colCosts</code>&nbsp;。</p>\n\n<ul>\n\t<li>如果机器人往 <strong>上</strong>&nbsp;或者往 <strong>下</strong>&nbsp;移动到第 <code>r</code>&nbsp;<strong>行</strong>&nbsp;的格子，那么代价为&nbsp;<code>rowCosts[r]</code>&nbsp;。</li>\n\t<li>如果机器人往 <strong>左</strong>&nbsp;或者往 <strong>右</strong>&nbsp;移动到第 <code>c</code>&nbsp;<strong>列</strong> 的格子，那么代价为&nbsp;<code>colCosts[c]</code>&nbsp;。</li>\n</ul>\n\n<p>请你返回机器人回家需要的 <strong>最小总代价</strong>&nbsp;。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<p><img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/10/11/eg-1.png\" style=\"width: 282px; height: 217px;\"></p>\n\n<pre><strong>输入：</strong>startPos = [1, 0], homePos = [2, 3], rowCosts = [5, 4, 3], colCosts = [8, 2, 6, 7]\n<b>输出：</b>18\n<b>解释：</b>一个最优路径为：\n从 (1, 0) 开始\n-&gt; 往下走到 (<em><strong>2</strong></em>, 0) 。代价为 rowCosts[2] = 3 。\n-&gt; 往右走到 (2, <em><strong>1</strong></em>) 。代价为 colCosts[1] = 2 。\n-&gt; 往右走到 (2, <em><strong>2</strong></em>) 。代价为 colCosts[2] = 6 。\n-&gt; 往右走到 (2, <em><strong>3</strong></em>) 。代价为 colCosts[3] = 7 。\n总代价为 3 + 2 + 6 + 7 = 18</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre><b>输入：</b>startPos = [0, 0], homePos = [0, 0], rowCosts = [5], colCosts = [26]\n<b>输出：</b>0\n<b>解释：</b>机器人已经在家了，所以不需要移动。总代价为 0 。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>m == rowCosts.length</code></li>\n\t<li><code>n == colCosts.length</code></li>\n\t<li><code>1 &lt;= m, n &lt;= 10<sup>5</sup></code></li>\n\t<li><code>0 &lt;= rowCosts[r], colCosts[c] &lt;= 10<sup>4</sup></code></li>\n\t<li><code>startPos.length == 2</code></li>\n\t<li><code>homePos.length == 2</code></li>\n\t<li><code>0 &lt;= start<sub>row</sub>, home<sub>row</sub> &lt; m</code></li>\n\t<li><code>0 &lt;= start<sub>col</sub>, home<sub>col</sub> &lt; n</code></li>\n</ul>\n",
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<code>use-after-free<\\/code> \\u9519\\u8bef\\u3002<\\/p>\\r\\n\\r\\n<p>\\u4e3a\\u4e86\\u4f7f\\u7528\\u65b9\\u4fbf\\uff0c\\u5927\\u90e8\\u5206\\u6807\\u51c6\\u5e93\\u7684\\u5934\\u6587\\u4ef6\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n<p>\\u5982\\u60f3\\u4f7f\\u7528\\u54c8\\u5e0c\\u8868\\u8fd0\\u7b97, \\u60a8\\u53ef\\u4ee5\\u4f7f\\u7528 <a href=\\\"https:\\/\\/troydhanson.github.io\\/uthash\\/\\\" target=\\\"_blank\\\">uthash<\\/a>\\u3002 \\\"uthash.h\\\"\\u5df2\\u7ecf\\u9ed8\\u8ba4\\u88ab\\u5bfc\\u5165\\u3002\\u8bf7\\u770b\\u5982\\u4e0b\\u793a\\u4f8b:<\\/p>\\r\\n\\r\\n<p><b>1. \\u5f80\\u54c8\\u5e0c\\u8868\\u4e2d\\u6dfb\\u52a0\\u4e00\\u4e2a\\u5bf9\\u8c61\\uff1a<\\/b>\\r\\n<pre>\\r\\nstruct hash_entry {\\r\\n    int id;            \\/* we'll use this field as the key *\\/\\r\\n    char name[10];\\r\\n    UT_hash_handle hh; \\/* makes this structure hashable *\\/\\r\\n};\\r\\n\\r\\nstruct hash_entry *users = NULL;\\r\\n\\r\\nvoid add_user(struct hash_entry 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5.1.6<\\/p>\\r\\n\\r\\n<p>Compile Options: --alwaysStrict --strictBindCallApply --strictFunctionTypes --target ES2022<\\/p>\\r\\n\\r\\n<p><a href=\\\"https:\\/\\/lodash.com\\\" target=\\\"_blank\\\">lodash.js<\\/a> \\u5e93\\u5df2\\u7ecf\\u9ed8\\u8ba4\\u88ab\\u5305\\u542b\\u3002<\\/p>\\r\\n\\r\\n<p> \\u5982\\u9700\\u4f7f\\u7528\\u961f\\u5217\\/\\u4f18\\u5148\\u961f\\u5217\\uff0c\\u60a8\\u53ef\\u4f7f\\u7528 <a href=\\\"https:\\/\\/github.com\\/datastructures-js\\/priority-queue\\/blob\\/v5\\/README.md\\\" target=\\\"_blank\\\"> datastructures-js\\/priority-queue@5.4.0<\\/a>\\uff0c<a href=\\\"https:\\/\\/github.com\\/datastructures-js\\/queue\\/tree\\/v4.2.3\\\" target=\\\"_blank\\\"> datastructures-js\\/queue@4.2.3<\\/a> \\u4ee5\\u53ca <a href=\\\"https:\\/\\/github.com\\/datastructures-js\\/deque\\/tree\\/v1.0.4\\\" target=\\\"_blank\\\"> datastructures-js\\/deque@1.0.4<\\/a>\\u3002<\\/p>\"],\"racket\":[\"Racket\",\"<p><a href=\\\"https:\\/\\/docs.racket-lang.org\\/guide\\/performance.html#%28tech._c%29\\\" target=\\\"_blank\\\">Racket CS<\\/a> v8.15<\\/p>\\r\\n\\r\\n<p>\\u4f7f\\u7528 #lang racket<\\/p>\\r\\n\\r\\n<p>\\u5df2\\u9884\\u5148 (require data\\/gvector data\\/queue data\\/order data\\/heap). \\u82e5\\u9700\\u4f7f\\u7528\\u5176\\u5b83\\u6570\\u636e\\u7ed3\\u6784\\uff0c\\u53ef\\u81ea\\u884c require\\u3002<\\/p>\"],\"erlang\":[\"Erlang\",\"Erlang\\/OTP 26\"],\"elixir\":[\"Elixir\",\"Elixir 1.17 with Erlang\\/OTP 26\"],\"dart\":[\"Dart\",\"<p>Dart 3.2\\u3002\\u60a8\\u53ef\\u4ee5\\u4f7f\\u7528 <a href=\\\"https:\\/\\/pub.dev\\/packages\\/collection\\/versions\\/1.18.0\\\" target=\\\"_blank\\\">collection<\\/a> \\u5305<\\/p>\\r\\n\\r\\n<p>\\u60a8\\u7684\\u4ee3\\u7801\\u5c06\\u4f1a\\u88ab\\u4e0d\\u7f16\\u8bd1\\u76f4\\u63a5\\u8fd0\\u884c<\\/p>\"],\"cangjie\":[\"Cangjie\",\"<p>\\u7248\\u672c\\uff1a0.53.11 (cjnative)<\\/p>\\r\\n\\r\\n<p>\\u7f16\\u8bd1\\u53c2\\u6570\\uff1a<code>-O2 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