leetcode-problemset/leetcode-cn/originData/vvXgSW.json

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        "question": {
            "questionId": "1000342",
            "questionFrontendId": "LCR 078",
            "categoryTitle": "Algorithms",
            "boundTopicId": 910336,
            "title": "合并 K 个升序链表",
            "titleSlug": "vvXgSW",
            "content": "<p>English description is not available for the problem. Please switch to Chinese.</p>\n",
            "translatedTitle": "合并 K 个升序链表",
            "translatedContent": "<p>给定一个链表数组，每个链表都已经按升序排列。</p>\n\n<p>请将所有链表合并到一个升序链表中，返回合并后的链表。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>lists = [[1,4,5],[1,3,4],[2,6]]\n<strong>输出：</strong>[1,1,2,3,4,4,5,6]\n<strong>解释：</strong>链表数组如下：\n[\n  1-&gt;4-&gt;5,\n  1-&gt;3-&gt;4,\n  2-&gt;6\n]\n将它们合并到一个有序链表中得到。\n1-&gt;1-&gt;2-&gt;3-&gt;4-&gt;4-&gt;5-&gt;6\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>lists = []\n<strong>输出：</strong>[]\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>lists = [[]]\n<strong>输出：</strong>[]\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>k == lists.length</code></li>\n\t<li><code>0 &lt;= k &lt;= 10^4</code></li>\n\t<li><code>0 &lt;= lists[i].length &lt;= 500</code></li>\n\t<li><code>-10^4 &lt;= lists[i][j] &lt;= 10^4</code></li>\n\t<li><code>lists[i]</code> 按 <strong>升序</strong> 排列</li>\n\t<li><code>lists[i].length</code> 的总和不超过 <code>10^4</code></li>\n</ul>\n\n<p>&nbsp;</p>\n\n<p><meta charset=\"UTF-8\" />注意：本题与主站 23&nbsp;题相同：&nbsp;<a href=\"https://leetcode-cn.com/problems/merge-k-sorted-lists/\">https://leetcode-cn.com/problems/merge-k-sorted-lists/</a></p>\n",
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                    "code": "/**\n * Definition for singly-linked list.\n * struct ListNode {\n *     int val;\n *     ListNode *next;\n *     ListNode() : val(0), next(nullptr) {}\n *     ListNode(int x) : val(x), next(nullptr) {}\n *     ListNode(int x, ListNode *next) : val(x), next(next) {}\n * };\n */\nclass Solution {\npublic:\n    ListNode* mergeKLists(vector<ListNode*>& lists) {\n\n    }\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 mergeKLists(self, lists: List[ListNode]) -> ListNode:",
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                    "code": "/**\n * Definition for singly-linked list.\n * struct ListNode {\n *     int val;\n *     struct ListNode *next;\n * };\n */\n\n\nstruct ListNode* mergeKLists(struct ListNode** lists, int listsSize){\n\n}",
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                    "code": "/**\n * Definition for singly-linked list.\n * function ListNode(val, next) {\n *     this.val = (val===undefined ? 0 : val)\n *     this.next = (next===undefined ? null : next)\n * }\n */\n/**\n * @param {ListNode[]} lists\n * @return {ListNode}\n */\nvar mergeKLists = function(lists) {\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 mergeKLists(lists: Array<ListNode | null>): ListNode | null {\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 mergeKLists(lists []*ListNode) *ListNode {\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_k_lists(lists: Vec<Option<Box<ListNode>>>) -> Option<Box<ListNode>> {\n\n    }\n}",
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