leetcode-problemset/leetcode-cn/originData/sum-lists-lcci.json

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        "question": {
            "questionId": "100188",
            "questionFrontendId": "面试题 02.05",
            "categoryTitle": "LCCI",
            "boundTopicId": 47067,
            "title": "Sum Lists LCCI",
            "titleSlug": "sum-lists-lcci",
            "content": "<p>You have two numbers represented by a linked list, where each node contains a single digit. The digits are stored in reverse order, such that the 1&#39;s digit is at the head of the list. Write a function that adds the two numbers and returns the sum as a linked list.</p>\r\n\r\n<p>&nbsp;</p>\r\n\r\n<p><strong>Example: </strong></p>\r\n\r\n<pre>\r\n<strong>Input: </strong>(7 -&gt; 1 -&gt; 6) + (5 -&gt; 9 -&gt; 2). That is, 617 + 295.\r\n<strong>Output: </strong>2 -&gt; 1 -&gt; 9. That is, 912.\r\n</pre>\r\n\r\n<p><strong>Follow Up:&nbsp;</strong>Suppose the digits are stored in forward order. Repeat the above problem.</p>\r\n\r\n<p><strong>Example: </strong></p>\r\n\r\n<pre>\r\n<strong>Input: </strong>(6 -&gt; 1 -&gt; 7) + (2 -&gt; 9 -&gt; 5). That is, 617 + 295.\r\n<strong>Output: </strong>9 -&gt; 1 -&gt; 2. That is, 912.\r\n</pre>\r\n",
            "translatedTitle": "链表求和",
            "translatedContent": "<p>给定两个用链表表示的整数，每个节点包含一个数位。</p>\n\n<p>这些数位是反向存放的，也就是个位排在链表首部。</p>\n\n<p>编写函数对这两个整数求和，并用链表形式返回结果。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例：</strong></p>\n\n<pre><strong>输入：</strong>(7 -&gt; 1 -&gt; 6) + (5 -&gt; 9 -&gt; 2)，即617 + 295\n<strong>输出：</strong>2 -&gt; 1 -&gt; 9，即912\n</pre>\n\n<p><strong>进阶：</strong>思考一下，假设这些数位是正向存放的，又该如何解决呢?</p>\n\n<p><strong>示例：</strong></p>\n\n<pre><strong>输入：</strong>(6 -&gt; 1 -&gt; 7) + (2 -&gt; 9 -&gt; 5)，即617 + 295\n<strong>输出：</strong>9 -&gt; 1 -&gt; 2，即912\n</pre>\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 addTwoNumbers(self, l1, l2):\n        \"\"\"\n        :type l1: ListNode\n        :type l2: ListNode\n        :rtype: ListNode\n        \"\"\"",
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                    "code": "/**\n * Definition for singly-linked list.\n * class ListNode {\n *     val: number\n *     next: ListNode | null\n *     constructor(val?: number, next?: ListNode | null) {\n *         this.val = (val===undefined ? 0 : val)\n *         this.next = (next===undefined ? null : next)\n *     }\n * }\n */\n\nfunction addTwoNumbers(l1: ListNode | null, l2: ListNode | null): ListNode | null {\n\n};",
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                    "code": "/**\n * Definition for a singly-linked list.\n * class ListNode {\n *     public $val = 0;\n *     public $next = null;\n *     function __construct($val) { $this->val = $val; }\n * }\n */\nclass Solution {\n\n    /**\n     * @param ListNode $l1\n     * @param ListNode $l2\n     * @return ListNode\n     */\n    function addTwoNumbers($l1, $l2) {\n\n    }\n}",
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                    "code": "/**\n * Definition for singly-linked list.\n * public class ListNode {\n *     public var val: Int\n *     public var next: ListNode?\n *     public init(_ val: Int) {\n *         self.val = val\n *         self.next = nil\n *     }\n * }\n */\nclass Solution {\n    func addTwoNumbers(_ l1: ListNode?, _ l2: ListNode?) -> ListNode? {\n\n    }\n}",
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                    "code": "/**\n * Example:\n * var li = ListNode(5)\n * var v = li.`val`\n * Definition for singly-linked list.\n * class ListNode(var `val`: Int) {\n *     var next: ListNode? = null\n * }\n */\nclass Solution {\n    fun addTwoNumbers(l1: ListNode?, l2: ListNode?): ListNode? {\n\n    }\n}",
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                    "code": "/**\n * Definition for singly-linked list.\n * class ListNode {\n *   int val;\n *   ListNode? next;\n *   ListNode([this.val = 0, this.next]);\n * }\n */\nclass Solution {\n  ListNode? addTwoNumbers(ListNode? l1, ListNode? l2) {\n\n  }\n}",
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                    "code": "/**\n * Definition for singly-linked list.\n * type ListNode struct {\n *     Val int\n *     Next *ListNode\n * }\n */\nfunc addTwoNumbers(l1 *ListNode, l2 *ListNode) *ListNode {\n\n}",
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                    "code": "/**\n * Definition for singly-linked list.\n * class ListNode(var _x: Int = 0) {\n *   var next: ListNode = null\n *   var x: Int = _x\n * }\n */\nobject Solution {\n    def addTwoNumbers(l1: ListNode, l2: ListNode): ListNode = {\n\n    }\n}",
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                    "code": "// Definition for singly-linked list.\n// #[derive(PartialEq, Eq, Clone, Debug)]\n// pub struct ListNode {\n//   pub val: i32,\n//   pub next: Option<Box<ListNode>>\n// }\n// \n// impl ListNode {\n//   #[inline]\n//   fn new(val: i32) -> Self {\n//     ListNode {\n//       next: None,\n//       val\n//     }\n//   }\n// }\nimpl Solution {\n    pub fn add_two_numbers(l1: Option<Box<ListNode>>, l2: Option<Box<ListNode>>) -> Option<Box<ListNode>> {\n\n    }\n}",
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                    "code": "; Definition for singly-linked list:\n#|\n\n; val : integer?\n; next : (or/c list-node? #f)\n(struct list-node\n  (val next) #:mutable #:transparent)\n\n; constructor\n(define (make-list-node [val 0])\n  (list-node val #f))\n\n|#\n\n(define/contract (add-two-numbers l1 l2)\n  (-> (or/c list-node? #f) (or/c list-node? #f) (or/c list-node? #f))\n\n  )",
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                    "code": "%% Definition for singly-linked list.\n%%\n%% -record(list_node, {val = 0 :: integer(),\n%%                     next = null :: 'null' | #list_node{}}).\n\n-spec add_two_numbers(L1 :: #list_node{} | null, L2 :: #list_node{} | null) -> #list_node{} | null.\nadd_two_numbers(L1, L2) ->\n  .",
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                    "code": "# Definition for singly-linked list.\n#\n# defmodule ListNode do\n#   @type t :: %__MODULE__{\n#           val: integer,\n#           next: ListNode.t() | nil\n#         }\n#   defstruct val: 0, next: nil\n# end\n\ndefmodule Solution do\n  @spec add_two_numbers(l1 :: ListNode.t | nil, l2 :: ListNode.t | nil) :: ListNode.t | nil\n  def add_two_numbers(l1, l2) do\n\n  end\nend",
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            "hints": [
                "当然，你可以将链表转换为整数，计算总和，然后将其转换回新的链表。如果你在面试中这样做，面试官可能会接受答案，然后看看你在不能将其转换为数字然后返回的情况下，还能否做到这一点。",
                "尝试递归。假设你有两个链表，A = 1 -> 5 -> 9（代表951）和B = 2 -> 3 -> 6 -> 7（代表7632），以及一个操作链表其余部分的函数（5 -> 9和3 -> 6 -> 7）。你能用这个来创建求和方法吗？sum(1 -> 5 -> 9, 2 -> 3 -> 6 -> 7)和sum(5 -> 9, 3 -> 6 -> 7)之间有何关系？",
                "确保你考虑到了链表的长度不同的情况。",
                "你的算法在形如9 -> 7 -> 8和6 -> 8 -> 5的链表上工作吗？仔细检查一下。",
                "对于后续问题：问题是，当链表的长度不一样时，一个链表的首部可能代表1000的位置，而另一个链表代表10的位置。如果你把它们做的一样长呢？有没有方法修改链表来做到这一点，而不改变它所代表的值？"
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