leetcode-problemset/leetcode-cn/originData/linked-list-cycle-lcci.json

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            "questionId": "100168",
            "questionFrontendId": "面试题 02.08",
            "categoryTitle": "LCCI",
            "boundTopicId": 45974,
            "title": "Linked List Cycle LCCI",
            "titleSlug": "linked-list-cycle-lcci",
            "content": "<p>Given a circular linked list, implement an algorithm that returns the node at the beginning of the loop.</p>\r\n\r\n<p>Circular linked list: A (corrupt) linked list in which a node&#39;s next pointer points to an earlier node, so as to make a loop in the linked list.</p>\r\n\r\n<p><strong>Example 1: </strong></p>\r\n\r\n<pre>\r\n<strong>Input: </strong>head = [3,2,0,-4], pos = 1\r\n<strong>Output: </strong>tail connects to node index 1</pre>\r\n\r\n<p><strong>Example 2: </strong></p>\r\n\r\n<pre>\r\n<strong>Input: </strong>head = [1,2], pos = 0\r\n<strong>Output: </strong>tail connects to node index 0</pre>\r\n\r\n<p><strong>Example 3: </strong></p>\r\n\r\n<pre>\r\n<strong>Input: </strong>head = [1], pos = -1\r\n<strong>Output: </strong>no cycle</pre>\r\n\r\n<p><strong>Follow Up: </strong><br />\r\nCan you solve it without using additional space?</p>\r\n",
            "translatedTitle": "环路检测",
            "translatedContent": "<p>给定一个链表，如果它是有环链表，实现一个算法返回环路的<code>开头节点</code>。若环不存在，请返回 <code>null</code>。</p>\n\n<p>如果链表中有某个节点，可以通过连续跟踪 <code>next</code> 指针再次到达，则链表中存在环。 为了表示给定链表中的环，我们使用整数 <code>pos</code> 来表示链表尾连接到链表中的位置（索引从 0 开始）。 如果 <code>pos</code> 是 <code>-1</code>，则在该链表中没有环。<strong>注意：<code>pos</code> 不作为参数进行传递</strong>，仅仅是为了标识链表的实际情况。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<p><img alt=\"\" src=\"https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/12/07/circularlinkedlist.png\" style=\"height: 97px; width: 300px;\" /></p>\n\n<pre>\n<strong>输入：</strong>head = [3,2,0,-4], pos = 1\n<strong>输出：</strong>tail connects to node index 1\n<strong>解释：</strong>链表中有一个环，其尾部连接到第二个节点。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<p><img alt=\"\" src=\"https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/12/07/circularlinkedlist_test2.png\" style=\"height: 74px; width: 141px;\" /></p>\n\n<pre>\n<strong>输入：</strong>head = [1,2], pos = 0\n<strong>输出：</strong>tail connects to node index 0\n<strong>解释：</strong>链表中有一个环，其尾部连接到第一个节点。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<p><img alt=\"\" src=\"https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/12/07/circularlinkedlist_test3.png\" style=\"height: 45px; width: 45px;\" /></p>\n\n<pre>\n<strong>输入：</strong>head = [1], pos = -1\n<strong>输出：</strong>no cycle\n<strong>解释：</strong>链表中没有环。</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>进阶：</strong></p>\n\n<ul>\n\t<li>你是否可以不用额外空间解决此题？</li>\n</ul>\n\n<p>&nbsp;</p>\n",
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            "hints": [
                "这个问题实际上可以分为两个部分。首先，检测链表是否有循环。第二，找出循环开始的位置。",
                "要确定是否有一个循环，请尝试“快行指针”方法。让一个指针比另一个指针快。",
                "你可以使用两个指针，一个指针移动速度是另一个指针的两倍。如果有环，两个指针会碰撞。它们将同时降落在同一地点。它们在哪里相遇？为什么呢？",
                "如果你还没有确定两个指针的起始位置，请尝试使用链表1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> ?，其中 ? 链接到另一个节点。试着让 ? 成为第一个节点（即9指向1，使得整个链表是一个循环）。然后让 ? 成为节点2，然后成为节点3，然后成为节点4。这一模式是什么？你能解释一下为什么会这样吗？"
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