mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
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180 lines
26 KiB
JSON
180 lines
26 KiB
JSON
{
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"questionId": "2903",
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"questionFrontendId": "2807",
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"categoryTitle": "Algorithms",
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"boundTopicId": 2372234,
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"title": "Insert Greatest Common Divisors in Linked List",
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"titleSlug": "insert-greatest-common-divisors-in-linked-list",
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"content": "<p>Given the head of a linked list <code>head</code>, in which each node contains an integer value.</p>\n\n<p>Between every pair of adjacent nodes, insert a new node with a value equal to the <strong>greatest common divisor</strong> of them.</p>\n\n<p>Return <em>the linked list after insertion</em>.</p>\n\n<p>The <strong>greatest common divisor</strong> of two numbers is the largest positive integer that evenly divides both numbers.</p>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2023/07/18/ex1_copy.png\" style=\"width: 641px; height: 181px;\" />\n<pre>\n<strong>Input:</strong> head = [18,6,10,3]\n<strong>Output:</strong> [18,6,6,2,10,1,3]\n<strong>Explanation:</strong> The 1<sup>st</sup> diagram denotes the initial linked list and the 2<sup>nd</sup> diagram denotes the linked list after inserting the new nodes (nodes in blue are the inserted nodes).\n- We insert the greatest common divisor of 18 and 6 = 6 between the 1<sup>st</sup> and the 2<sup>nd</sup> nodes.\n- We insert the greatest common divisor of 6 and 10 = 2 between the 2<sup>nd</sup> and the 3<sup>rd</sup> nodes.\n- We insert the greatest common divisor of 10 and 3 = 1 between the 3<sup>rd</sup> and the 4<sup>th</sup> nodes.\nThere are no more adjacent nodes, so we return the linked list.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2023/07/18/ex2_copy1.png\" style=\"width: 51px; height: 191px;\" />\n<pre>\n<strong>Input:</strong> head = [7]\n<strong>Output:</strong> [7]\n<strong>Explanation:</strong> The 1<sup>st</sup> diagram denotes the initial linked list and the 2<sup>nd</sup> diagram denotes the linked list after inserting the new nodes.\nThere are no pairs of adjacent nodes, so we return the initial linked list.\n</pre>\n\n<p> </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>[1, 5000]</code>.</li>\n\t<li><code>1 <= Node.val <= 1000</code></li>\n</ul>\n",
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"translatedTitle": "在链表中插入最大公约数",
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"translatedContent": "<p>给你一个链表的头 <code>head</code> ,每个结点包含一个整数值。</p>\n\n<p>在相邻结点之间,请你插入一个新的结点,结点值为这两个相邻结点值的 <strong>最大公约数</strong> 。</p>\n\n<p>请你返回插入之后的链表。</p>\n\n<p>两个数的 <strong>最大公约数</strong> 是可以被两个数字整除的最大正整数。</p>\n\n<p> </p>\n\n<p><strong>示例 1:</strong></p>\n\n<p><img alt=\"\" src=\"https://assets.leetcode.com/uploads/2023/07/18/ex1_copy.png\" style=\"width: 641px; height: 181px;\"></p>\n\n<pre><b>输入:</b>head = [18,6,10,3]\n<b>输出:</b>[18,6,6,2,10,1,3]\n<b>解释:</b>第一幅图是一开始的链表,第二幅图是插入新结点后的图(蓝色结点为新插入结点)。\n- 18 和 6 的最大公约数为 6 ,插入第一和第二个结点之间。\n- 6 和 10 的最大公约数为 2 ,插入第二和第三个结点之间。\n- 10 和 3 的最大公约数为 1 ,插入第三和第四个结点之间。\n所有相邻结点之间都插入完毕,返回链表。\n</pre>\n\n<p><strong>示例 2:</strong></p>\n\n<p><img alt=\"\" src=\"https://assets.leetcode.com/uploads/2023/07/18/ex2_copy1.png\" style=\"width: 51px; height: 191px;\"></p>\n\n<pre><b>输入:</b>head = [7]\n<strong>输出:</strong>[7]\n<b>解释:</b>第一幅图是一开始的链表,第二幅图是插入新结点后的图(蓝色结点为新插入结点)。\n没有相邻结点,所以返回初始链表。\n</pre>\n\n<p> </p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li>链表中结点数目在 <code>[1, 5000]</code> 之间。</li>\n\t<li><code>1 <= Node.val <= 1000</code></li>\n</ul>\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? insertGreatestCommonDivisors(ListNode? head) {\n\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 insert_greatest_common_divisors(head)\n\nend",
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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 insert_greatest_common_divisors(Head :: #list_node{} | null) -> #list_node{} | null.\ninsert_greatest_common_divisors(Head) ->\n .",
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