leetcode-problemset/leetcode-cn/originData/intersection-of-two-linked-lists-lcci.json

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        "question": {
            "questionId": "100167",
            "questionFrontendId": "面试题 02.07",
            "categoryTitle": "LCCI",
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            "title": "Intersection of Two Linked Lists LCCI",
            "titleSlug": "intersection-of-two-linked-lists-lcci",
            "content": "<p>Given two (singly) linked lists, determine if the two lists intersect. Return the inter&shy; secting node. Note that the intersection is defined based on reference, not value. That is, if the kth node of the first linked list is the exact same node (by reference) as the jth node of the second linked list, then they are intersecting.</p>\r\n\r\n<p><strong>Example 1: </strong></p>\r\n\r\n<pre>\r\n<strong>Input: </strong>intersectVal = 8, listA = [4,1,8,4,5], listB = [5,0,1,8,4,5], skipA = 2, skipB = 3\r\n<strong>Output: </strong>Reference of the node with value = 8\r\n<strong>Input Explanation:</strong> The intersected node&#39;s value is 8 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [4,1,8,4,5]. From the head of B, it reads as [5,0,1,8,4,5]. There are 2 nodes before the intersected node in A; There are 3 nodes before the intersected node in B.</pre>\r\n\r\n<p><strong>Example 2: </strong></p>\r\n\r\n<pre>\r\n<strong>Input: </strong>intersectVal = 2, listA = [0,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1\r\n<strong>Output: </strong>Reference of the node with value = 2\r\n<strong>Input Explanation:</strong>&nbsp;The intersected node&#39;s value is 2 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [0,9,1,2,4]. From the head of B, it reads as [3,2,4]. There are 3 nodes before the intersected node in A; There are 1 node before the intersected node in B.</pre>\r\n\r\n<p><strong>Example 3: </strong></p>\r\n\r\n<pre>\r\n<strong>Input: </strong>intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2\r\n<strong>Output: </strong>null\r\n<strong>Input Explanation:</strong> From the head of A, it reads as [2,6,4]. From the head of B, it reads as [1,5]. Since the two lists do not intersect, intersectVal must be 0, while skipA and skipB can be arbitrary values.\r\n<strong>Explanation:</strong> The two lists do not intersect, so return null.</pre>\r\n\r\n<p><b>Notes:</b></p>\r\n\r\n<ul>\r\n\t<li>If the two linked lists have no intersection at all, return&nbsp;<code>null</code>.</li>\r\n\t<li>The linked lists must retain their original structure after the function returns.</li>\r\n\t<li>You may assume there are no cycles anywhere in the entire linked structure.</li>\r\n\t<li>Your code should preferably run in O(n) time and use only O(1) memory.</li>\r\n</ul>\r\n",
            "translatedTitle": "链表相交",
            "translatedContent": "<p>给你两个单链表的头节点 <code>headA</code> 和 <code>headB</code> ，请你找出并返回两个单链表相交的起始节点。如果两个链表没有交点，返回 <code>null</code> 。</p>\n\n<p>图示两个链表在节点 <code>c1</code> 开始相交<strong>：</strong></p>\n\n<p><a href=\"https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/12/14/160_statement.png\" target=\"_blank\"><img alt=\"\" src=\"https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/12/14/160_statement.png\" style=\"height: 130px; width: 400px;\" /></a></p>\n\n<p>题目数据 <strong>保证</strong> 整个链式结构中不存在环。</p>\n\n<p><strong>注意</strong>，函数返回结果后，链表必须 <strong>保持其原始结构</strong> 。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<p><a href=\"https://assets.leetcode.com/uploads/2018/12/13/160_example_1.png\" target=\"_blank\"><img alt=\"\" src=\"https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/12/14/160_example_1.png\" style=\"height: 130px; width: 400px;\" /></a></p>\n\n<pre>\n<strong>输入：</strong>intersectVal = 8, listA = [4,1,8,4,5], listB = [5,0,1,8,4,5], skipA = 2, skipB = 3\n<strong>输出：</strong>Intersected at '8'\n<strong>解释：</strong>相交节点的值为 8 （注意，如果两个链表相交则不能为 0）。\n从各自的表头开始算起，链表 A 为 [4,1,8,4,5]，链表 B 为 [5,0,1,8,4,5]。\n在 A 中，相交节点前有 2 个节点；在 B 中，相交节点前有 3 个节点。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<p><a href=\"https://assets.leetcode.com/uploads/2018/12/13/160_example_2.png\" target=\"_blank\"><img alt=\"\" src=\"https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/12/14/160_example_2.png\" style=\"height: 136px; width: 350px;\" /></a></p>\n\n<pre>\n<strong>输入：</strong>intersectVal = 2, listA = [0,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1\n<strong>输出：</strong>Intersected at '2'\n<strong>解释：</strong>相交节点的值为 2 （注意，如果两个链表相交则不能为 0）。\n从各自的表头开始算起，链表 A 为 [0,9,1,2,4]，链表 B 为 [3,2,4]。\n在 A 中，相交节点前有 3 个节点；在 B 中，相交节点前有 1 个节点。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<p><a href=\"https://assets.leetcode.com/uploads/2018/12/13/160_example_3.png\" target=\"_blank\"><img alt=\"\" src=\"https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/12/14/160_example_3.png\" style=\"height: 126px; width: 200px;\" /></a></p>\n\n<pre>\n<strong>输入：</strong>intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2\n<strong>输出：</strong>null\n<strong>解释：</strong>从各自的表头开始算起，链表 A 为 [2,6,4]，链表 B 为 [1,5]。\n由于这两个链表不相交，所以 intersectVal 必须为 0，而 skipA 和 skipB 可以是任意值。\n这两个链表不相交，因此返回 null 。\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>listA</code> 中节点数目为 <code>m</code></li>\n\t<li><code>listB</code> 中节点数目为 <code>n</code></li>\n\t<li><code>0 <= m, n <= 3 * 10<sup>4</sup></code></li>\n\t<li><code>1 <= Node.val <= 10<sup>5</sup></code></li>\n\t<li><code>0 <= skipA <= m</code></li>\n\t<li><code>0 <= skipB <= n</code></li>\n\t<li>如果 <code>listA</code> 和 <code>listB</code> 没有交点，<code>intersectVal</code> 为 <code>0</code></li>\n\t<li>如果 <code>listA</code> 和 <code>listB</code> 有交点，<code>intersectVal == listA[skipA + 1] == listB[skipB + 1]</code></li>\n</ul>\n\n<p> </p>\n\n<p><strong>进阶：</strong>你能否设计一个时间复杂度 <code>O(n)</code> 、仅用 <code>O(1)</code> 内存的解决方案？</p>\n",
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                    "code": "# Definition for singly-linked list.\n# class ListNode(object):\n#     def __init__(self, x):\n#         self.val = x\n#         self.next = None\n\nclass Solution(object):\n    def getIntersectionNode(self, headA, headB):\n        \"\"\"\n        :type head1, head1: ListNode\n        :rtype: ListNode\n        \"\"\"\n        ",
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                    "code": "# Definition for singly-linked list.\n# class ListNode:\n#     def __init__(self, x):\n#         self.val = x\n#         self.next = None\n\nclass Solution:\n    def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode:\n        ",
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            "hints": [
                "你可以在O(A + B)的时间和额外的O(1)空间中做到这一点，也就是说，你不需要一个散列表（尽管你可以用一个散列表来完成）。",
                "举例子能帮到你。画一个相交的链表和两个不相交的等价链表（值）的图片。",
                "首先要确定是否有交叉点。",
                "注意，两个相交链表的最后节点始终相同。一旦它们相交，之后的所有节点将相等。",
                "你可以通过遍历到每个链表的末尾并比较它们的尾节点来确定两个链表是否相交。",
                "现在，你需要查找链表在何处相交。假设链表长度相同。你可以怎么做？",
                "如果两个链表长度相同，则可以在每个链表中向前遍历，直到找到一个公共的元素。现在，面对长度不同的链表，你该怎样调整？",
                "尝试使用两个链表长度之间的差异。",
                "如果你通过长度差异向较长的链表中移动指针，则可以在链表相同时应用类似的方法。"
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