leetcode-problemset/leetcode-cn/originData/merge-nodes-in-between-zeros.json

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        "question": {
            "questionId": "2299",
            "questionFrontendId": "2181",
            "categoryTitle": "Algorithms",
            "boundTopicId": 1274498,
            "title": "Merge Nodes in Between Zeros",
            "titleSlug": "merge-nodes-in-between-zeros",
            "content": "<p>You are given the <code>head</code> of a linked list, which contains a series of integers <strong>separated</strong> by <code>0</code>&#39;s. The <strong>beginning</strong> and <strong>end</strong> of the linked list will have <code>Node.val == 0</code>.</p>\n\n<p>For <strong>every </strong>two consecutive <code>0</code>&#39;s, <strong>merge</strong> all the nodes lying in between them into a single node whose value is the <strong>sum</strong> of all the merged nodes. The modified list should not contain any <code>0</code>&#39;s.</p>\n\n<p>Return <em>the</em> <code>head</code> <em>of the modified linked list</em>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2022/02/02/ex1-1.png\" style=\"width: 600px; height: 41px;\" />\n<pre>\n<strong>Input:</strong> head = [0,3,1,0,4,5,2,0]\n<strong>Output:</strong> [4,11]\n<strong>Explanation:</strong> \nThe above figure represents the given linked list. The modified list contains\n- The sum of the nodes marked in green: 3 + 1 = 4.\n- The sum of the nodes marked in red: 4 + 5 + 2 = 11.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2022/02/02/ex2-1.png\" style=\"width: 600px; height: 41px;\" />\n<pre>\n<strong>Input:</strong> head = [0,1,0,3,0,2,2,0]\n<strong>Output:</strong> [1,3,4]\n<strong>Explanation:</strong> \nThe above figure represents the given linked list. The modified list contains\n- The sum of the nodes marked in green: 1 = 1.\n- The sum of the nodes marked in red: 3 = 3.\n- The sum of the nodes marked in yellow: 2 + 2 = 4.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li>The number of nodes in the list is in the range <code>[3, 2 * 10<sup>5</sup>]</code>.</li>\n\t<li><code>0 &lt;= Node.val &lt;= 1000</code></li>\n\t<li>There are <strong>no</strong> two consecutive nodes with <code>Node.val == 0</code>.</li>\n\t<li>The <strong>beginning</strong> and <strong>end</strong> of the linked list have <code>Node.val == 0</code>.</li>\n</ul>\n",
            "translatedTitle": "合并零之间的节点",
            "translatedContent": "<p>给你一个链表的头节点 <code>head</code> ，该链表包含由 <code>0</code> 分隔开的一连串整数。链表的 <strong>开端</strong> 和 <strong>末尾</strong> 的节点都满足 <code>Node.val == 0</code> 。</p>\n\n<p>对于每两个相邻的 <code>0</code> ，请你将它们之间的所有节点合并成一个节点，其值是所有已合并节点的值之和。然后将所有 <code>0</code> 移除，修改后的链表不应该含有任何 <code>0</code> 。</p>\n\n<p>&nbsp;返回修改后链表的头节点 <code>head</code> 。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：<br />\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2022/02/02/ex1-1.png\" style=\"width: 600px; height: 41px;\" /></strong></p>\n\n<pre>\n<strong>输入：</strong>head = [0,3,1,0,4,5,2,0]\n<strong>输出：</strong>[4,11]\n<strong>解释：</strong>\n上图表示输入的链表。修改后的链表包含：\n- 标记为绿色的节点之和：3 + 1 = 4\n- 标记为红色的节点之和：4 + 5 + 2 = 11\n</pre>\n\n<p><strong>示例 2：<br />\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2022/02/02/ex2-1.png\" style=\"width: 600px; height: 41px;\" /></strong></p>\n\n<pre>\n<strong>输入：</strong>head = [0,1,0,3,0,2,2,0]\n<strong>输出：</strong>[1,3,4]\n<strong>解释：</strong>\n上图表示输入的链表。修改后的链表包含：\n- 标记为绿色的节点之和：1 = 1\n- 标记为红色的节点之和：3 = 3\n- 标记为黄色的节点之和：2 + 2 = 4\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li>列表中的节点数目在范围 <code>[3, 2 * 10<sup>5</sup>]</code> 内</li>\n\t<li><code>0 &lt;= Node.val &lt;= 1000</code></li>\n\t<li><strong>不</strong> 存在连续两个&nbsp;<code>Node.val == 0</code> 的节点</li>\n\t<li>链表的 <strong>开端</strong> 和 <strong>末尾</strong> 节点都满足 <code>Node.val == 0</code></li>\n</ul>\n",
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                    "code": "# Definition for singly-linked list.\n# class ListNode:\n#     def __init__(self, val=0, next=None):\n#         self.val = val\n#         self.next = next\nclass Solution:\n    def mergeNodes(self, head: Optional[ListNode]) -> Optional[ListNode]:\n        ",
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                    "code": "/**\n * Definition for singly-linked list.\n * public class ListNode {\n *     public int val;\n *     public ListNode next;\n *     public ListNode(int val=0, ListNode next=null) {\n *         this.val = val;\n *         this.next = next;\n *     }\n * }\n */\npublic class Solution {\n    public ListNode MergeNodes(ListNode head) {\n        \n    }\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 mergeNodes(head: 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 = 0, $next = null) {\n *         $this->val = $val;\n *         $this->next = $next;\n *     }\n * }\n */\nclass Solution {\n\n    /**\n     * @param ListNode $head\n     * @return ListNode\n     */\n    function mergeNodes($head) {\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 mergeNodes(head *ListNode) *ListNode {\n    \n}",
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                    "code": "# Definition for singly-linked list.\n# class ListNode\n#     attr_accessor :val, :next\n#     def initialize(val = 0, _next = nil)\n#         @val = val\n#         @next = _next\n#     end\n# end\n# @param {ListNode} head\n# @return {ListNode}\ndef merge_nodes(head)\n    \nend",
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                    "code": "/**\n * Definition for singly-linked list.\n * class ListNode(_x: Int = 0, _next: ListNode = null) {\n *   var next: ListNode = _next\n *   var x: Int = _x\n * }\n */\nobject Solution {\n    def mergeNodes(head: 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 merge_nodes(head: Option<Box<ListNode>>) -> Option<Box<ListNode>> {\n        \n    }\n}",
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                    "lang": "Racket",
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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 (merge-nodes head)\n  (-> (or/c list-node? #f) (or/c list-node? #f))\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 merge_nodes(Head :: #list_node{} | null) -> #list_node{} | null.\nmerge_nodes(Head) ->\n  .",
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                    "code": "/**\n * Definition for singly-linked list.\n * public class ListNode {\n *     public var val: Int64\n *     public var next: ?ListNode\n *     public init() {\n *         val = 0\n *         next = None\n *     }\n *     public init(val: Int64) {\n *         this.val = val\n *         next = None\n *     }\n * }\n */\n\nclass Solution {\n    func mergeNodes(head: ?ListNode): ?ListNode {\n\n    }\n}",
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            "hints": [
                "How can you use two pointers to modify the original list into the new list?",
                "Have a pointer traverse the entire linked list, while another pointer looks at a node that is currently being modified.",
                "Keep on summing the values of the nodes between the traversal pointer and the modifying pointer until the former comes across a ‘0’. In that case, the modifying pointer is incremented to modify the next node.",
                "Do not forget to have the next pointer of the final node of the modified list point to null."
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