leetcode-problemset/leetcode-cn/originData/you-le-yuan-de-mi-gong.json

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        "question": {
            "questionId": "1000088",
            "questionFrontendId": "LCP 15",
            "categoryTitle": "Algorithms",
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            "title": "游乐园的迷宫",
            "titleSlug": "you-le-yuan-de-mi-gong",
            "content": "English description is not available for the problem. Please switch to Chinese.",
            "translatedTitle": "游乐园的迷宫",
            "translatedContent": "<p>小王来到了游乐园，她玩的第一个项目是模拟推销员。有一个二维平面地图，其中散布着 <code>N</code> 个推销点，编号 <code>0</code> 到 <code>N-1</code>，不存在三点共线的情况。每两点之间有一条直线相连。游戏没有规定起点和终点，但限定了每次转角的方向。首先，小王需要先选择两个点分别作为起点和终点，然后从起点开始访问剩余 <code>N-2</code> 个点恰好一次并回到终点。访问的顺序需要满足一串给定的长度为 <code>N-2</code> 由 <code>L</code> 和 <code>R</code> 组成的字符串 <code>direction</code>，表示从起点出发之后在每个顶点上转角的方向。根据这个提示，小王希望你能够帮她找到一个可行的遍历顺序，输出顺序下标（若有多个方案，输出任意一种）。可以证明这样的遍历顺序一定是存在的。</p>\n\n<p><img alt=\"Screenshot 2020-03-20 at 17.04.58.png\" src=\"https://pic.leetcode-cn.com/595b60797d4a461287864a8cd05bba1d3b8760104ff83f43b902fd68477be9c3-Screenshot%202020-03-20%20at%2017.04.58.png\" style=\"max-height:500px\" /></p>\n\n<p>（上图：A-&gt;B-&gt;C 右转； 下图：D-&gt;E-&gt;F 左转）</p>\n\n<p><strong>示例 1：</strong></p>\n\n<blockquote>\n<p>输入：<code>points = [[1,1],[1,4],[3,2],[2,1]], direction = \"LL\"</code></p>\n\n<p>输入：<code>[0,2,1,3]</code></p>\n\n<p>解释：[0,2,1,3] 是符合\"LL\"的方案之一。在 [0,2,1,3] 方案中，0-&gt;2-&gt;1 是左转方向， 2-&gt;1-&gt;3 也是左转方向 <img alt=\"图片.gif\" src=\"https://pic.leetcode-cn.com/c01c1efc423b916267c2a3a170266c925c368d62afa047c267cc1020970e55d9-%E5%9B%BE%E7%89%87.gif\" style=\"max-height:300px\" /></p>\n</blockquote>\n\n<p><strong>示例 2：</strong></p>\n\n<blockquote>\n<p>输入：<code>points = [[1,3],[2,4],[3,3],[2,1]], direction = \"LR\"</code></p>\n\n<p>输入：<code>[0,3,1,2]</code></p>\n\n<p>解释：[0,3,1,2] 是符合\"LR\"的方案之一。在 [0,3,1,2] 方案中，0-&gt;3-&gt;1 是左转方向， 3-&gt;1-&gt;2 是右转方向</p>\n</blockquote>\n\n<p><strong>限制：</strong></p>\n\n<ul>\n\t<li><code>3 &lt;= points.length &lt;= 1000 且 points[i].length == 2</code></li>\n\t<li><code>1 &lt;= points[i][0],points[i][1] &lt;= 10000</code></li>\n\t<li><code>direction.length == points.length - 2</code></li>\n\t<li><code>direction 只包含 \"L\",\"R\"</code></li>\n</ul>\n",
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