![\"\"](\"https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/12/14/160_statement.png\")
Given two (singly) linked lists, determine if the two lists intersect. Return the inter 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.
\r\n\r\nExample 1:
\r\n\r\n\r\nInput: intersectVal = 8, listA = [4,1,8,4,5], listB = [5,0,1,8,4,5], skipA = 2, skipB = 3\r\nOutput: Reference of the node with value = 8\r\nInput Explanation: The intersected node'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.\r\n\r\n
Example 2:
\r\n\r\n\r\nInput: intersectVal = 2, listA = [0,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1\r\nOutput: Reference of the node with value = 2\r\nInput Explanation: The intersected node'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.\r\n\r\n
Example 3:
\r\n\r\n\r\nInput: intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2\r\nOutput: null\r\nInput Explanation: 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\nExplanation: The two lists do not intersect, so return null.\r\n\r\n
Notes:
\r\n\r\nnull.给你两个单链表的头节点 headA 和 headB ，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 null 。
图示两个链表在节点 c1 开始相交：
题目数据 保证 整个链式结构中不存在环。
\n\n注意，函数返回结果后，链表必须 保持其原始结构 。
\n\n\n\n
示例 1：
\n\n![\"\"](\"https://assets.leetcode.com/uploads/2018/12/13/160_example_1.png\")
\n输入：intersectVal = 8, listA = [4,1,8,4,5], listB = [5,0,1,8,4,5], skipA = 2, skipB = 3\n输出：Intersected at '8'\n解释：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\n\n
示例 2：
\n\n![\"\"](\"https://assets.leetcode.com/uploads/2018/12/13/160_example_2.png\")
\n输入：intersectVal = 2, listA = [0,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1\n输出：Intersected at '2'\n解释：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\n\n
示例 3：
\n\n![\"\"](\"https://assets.leetcode.com/uploads/2018/12/13/160_example_3.png\")
\n输入：intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2\n输出：null\n解释：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\n\n
\n\n
提示：
\n\nlistA 中节点数目为 mlistB 中节点数目为 n0 <= m, n <= 3 * 1041 <= Node.val <= 1050 <= skipA <= m0 <= skipB <= nlistA 和 listB 没有交点，intersectVal 为 0listA 和 listB 有交点，intersectVal == listA[skipA + 1] == listB[skipB + 1]\n\n
进阶：你能否设计一个时间复杂度 O(n) 、仅用 O(1) 内存的解决方案？
\\u7248\\u672c\\uff1a \\u7f16\\u8bd1\\u65f6\\uff0c\\u5c06\\u4f1a\\u91c7\\u7528 \\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>\"],\"java\":[\"Java\",\" \\u7248\\u672c\\uff1a \\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u6807\\u51c6\\u5e93\\u7684\\u5934\\u6587\\u4ef6\\u5df2\\u88ab\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n \\u5305\\u542b Pair \\u7c7b: https:\\/\\/docs.oracle.com\\/javase\\/8\\/javafx\\/api\\/javafx\\/util\\/Pair.html <\\/p>\"],\"python\":[\"Python\",\" \\u7248\\u672c\\uff1a \\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u5e38\\u7528\\u5e93\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8 \\u5bfc\\u5165\\uff0c\\u5982\\uff1aarray<\\/a>, bisect<\\/a>, collections<\\/a>\\u3002\\u5982\\u679c\\u60a8\\u9700\\u8981\\u4f7f\\u7528\\u5176\\u4ed6\\u5e93\\u51fd\\u6570\\uff0c\\u8bf7\\u81ea\\u884c\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n \\u6ce8\\u610f Python 2.7 \\u5c06\\u57282020\\u5e74\\u540e\\u4e0d\\u518d\\u7ef4\\u62a4<\\/a>\\u3002 \\u5982\\u60f3\\u4f7f\\u7528\\u6700\\u65b0\\u7248\\u7684Python\\uff0c\\u8bf7\\u9009\\u62e9Python 3\\u3002<\\/p>\"],\"c\":[\"C\",\" \\u7248\\u672c\\uff1a \\u7f16\\u8bd1\\u65f6\\uff0c\\u5c06\\u4f1a\\u91c7\\u7528 \\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 \\u5982\\u60f3\\u4f7f\\u7528\\u54c8\\u5e0c\\u8868\\u8fd0\\u7b97, \\u60a8\\u53ef\\u4ee5\\u4f7f\\u7528 uthash<\\/a>\\u3002 \\\"uthash.h\\\"\\u5df2\\u7ecf\\u9ed8\\u8ba4\\u88ab\\u5bfc\\u5165\\u3002\\u8bf7\\u770b\\u5982\\u4e0b\\u793a\\u4f8b:<\\/p>\\r\\n\\r\\n 1. \\u5f80\\u54c8\\u5e0c\\u8868\\u4e2d\\u6dfb\\u52a0\\u4e00\\u4e2a\\u5bf9\\u8c61\\uff1a<\\/b>\\r\\n 2. \\u5728\\u54c8\\u5e0c\\u8868\\u4e2d\\u67e5\\u627e\\u4e00\\u4e2a\\u5bf9\\u8c61\\uff1a<\\/b>\\r\\n 3. \\u4ece\\u54c8\\u5e0c\\u8868\\u4e2d\\u5220\\u9664\\u4e00\\u4e2a\\u5bf9\\u8c61\\uff1a<\\/b>\\r\\n C# 10<\\/a> \\u8fd0\\u884c\\u5728 .NET 6 \\u4e0a<\\/p>\\r\\n\\r\\n \\u60a8\\u7684\\u4ee3\\u7801\\u5728\\u7f16\\u8bd1\\u65f6\\u9ed8\\u8ba4\\u5f00\\u542f\\u4e86debug\\u6807\\u8bb0( \\u7248\\u672c\\uff1a \\u60a8\\u7684\\u4ee3\\u7801\\u5728\\u6267\\u884c\\u65f6\\u5c06\\u5e26\\u4e0a lodash.js<\\/a> \\u5e93\\u5df2\\u7ecf\\u9ed8\\u8ba4\\u88ab\\u5305\\u542b\\u3002<\\/p>\\r\\n\\r\\n \\u5982\\u9700\\u4f7f\\u7528\\u961f\\u5217\\/\\u4f18\\u5148\\u961f\\u5217\\uff0c\\u60a8\\u53ef\\u4f7f\\u7528 datastructures-js\\/priority-queue<\\/a> \\u548c datastructures-js\\/queue<\\/a>\\u3002<\\/p>\"],\"golang\":[\"Go\",\" \\u7248\\u672c\\uff1a \\u652f\\u6301 https:\\/\\/godoc.org\\/github.com\\/emirpasic\\/gods<\\/a> \\u7b2c\\u4e09\\u65b9\\u5e93\\u3002<\\/p>\"],\"python3\":[\"Python3\",\" \\u7248\\u672c\\uff1a \\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u5e38\\u7528\\u5e93\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8 \\u5bfc\\u5165\\uff0c\\u5982array<\\/a>, bisect<\\/a>, collections<\\/a>\\u3002 \\u5982\\u679c\\u60a8\\u9700\\u8981\\u4f7f\\u7528\\u5176\\u4ed6\\u5e93\\u51fd\\u6570\\uff0c\\u8bf7\\u81ea\\u884c\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n \\u5982\\u9700\\u4f7f\\u7528 Map\\/TreeMap \\u6570\\u636e\\u7ed3\\u6784\\uff0c\\u60a8\\u53ef\\u4f7f\\u7528 sortedcontainers<\\/a> \\u5e93\\u3002<\\/p>\"]}",
"book": {
"id": "2",
"bookName": "《程序员面试金典（第 6 版）》",
"pressName": "图灵公司",
"source": "《程序员面试金典（第 6 版）》独家授权",
"shortDescription": "英文版：Cracking the Coding Interview，源自硅谷程序员的面试红宝书",
"fullDescription": "本书是原谷歌资深面试官的经验之作，帮助了许多想要加入脸书、苹果、谷歌等 IT 名企的求职者拿到 Dream offer。本专题的 100+ 编程面试题是在原书基础上精心挑选出来的，帮助你轻松应战 IT 名企技术面试。",
"bookImgUrl": "https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2019/09/30/book_2.jpg",
"pressImgUrl": "https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2020/02/11/logo20190828-02.png",
"productUrl": "",
"__typename": "CategoryInfoNode"
},
"isSubscribed": false,
"isDailyQuestion": false,
"dailyRecordStatus": null,
"editorType": "CKEDITOR",
"ugcQuestionId": null,
"style": "LEETCODE",
"exampleTestcases": "8\n[4,1,8,4,5]\n[5,0,1,8,4,5]\n2\n3\n2\n[0,9,1,2,4]\n[3,2,4]\n3\n1\n0\n[2,6,4]\n[1,5]\n3\n2",
"__typename": "QuestionNode"
}
}
}clang 11<\\/code> \\u91c7\\u7528\\u6700\\u65b0C++ 17\\u6807\\u51c6\\u3002<\\/p>\\r\\n\\r\\n-O2<\\/code>\\u7ea7\\u4f18\\u5316\\u3002AddressSanitizer<\\/a> \\u4e5f\\u88ab\\u5f00\\u542f\\u6765\\u68c0\\u6d4bout-of-bounds<\\/code>\\u548cuse-after-free<\\/code>\\u9519\\u8bef\\u3002<\\/p>\\r\\n\\r\\nOpenJDK 17<\\/code>\\u3002\\u53ef\\u4ee5\\u4f7f\\u7528Java 8\\u7684\\u7279\\u6027\\u4f8b\\u5982\\uff0clambda expressions \\u548c stream API\\u3002<\\/p>\\r\\n\\r\\nPython 2.7.12<\\/code><\\/p>\\r\\n\\r\\nGCC 8.2<\\/code>\\uff0c\\u91c7\\u7528GNU99\\u6807\\u51c6\\u3002<\\/p>\\r\\n\\r\\n-O1<\\/code>\\u7ea7\\u4f18\\u5316\\u3002 AddressSanitizer<\\/a>\\u4e5f\\u88ab\\u5f00\\u542f\\u6765\\u68c0\\u6d4bout-of-bounds<\\/code>\\u548cuse-after-free<\\/code>\\u9519\\u8bef\\u3002<\\/p>\\r\\n\\r\\n\\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 *s) {\\r\\n HASH_ADD_INT(users, id, s);\\r\\n}\\r\\n<\\/pre>\\r\\n<\\/p>\\r\\n\\r\\n\\r\\nstruct hash_entry *find_user(int user_id) {\\r\\n struct hash_entry *s;\\r\\n HASH_FIND_INT(users, &user_id, s);\\r\\n return s;\\r\\n}\\r\\n<\\/pre>\\r\\n<\\/p>\\r\\n\\r\\n\\r\\nvoid delete_user(struct hash_entry *user) {\\r\\n HASH_DEL(users, user); \\r\\n}\\r\\n<\\/pre>\\r\\n<\\/p>\"],\"csharp\":[\"C#\",\"\\/debug:pdbonly<\\/code>)\\u3002<\\/p>\"],\"javascript\":[\"JavaScript\",\"Node.js 16.13.2<\\/code><\\/p>\\r\\n\\r\\n--harmony<\\/code> \\u6807\\u8bb0\\u6765\\u5f00\\u542f \\u65b0\\u7248ES6\\u7279\\u6027<\\/a>\\u3002<\\/p>\\r\\n\\r\\nGo 1.17<\\/code><\\/p>\\r\\n\\r\\nPython 3.10<\\/code><\\/p>\\r\\n\\r\\n