leetcode-problemset/leetcode-cn/originData/maximum-twin-sum-of-a-linked-list.json

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        "question": {
            "questionId": "2236",
            "questionFrontendId": "2130",
            "categoryTitle": "Algorithms",
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            "title": "Maximum Twin Sum of a Linked List",
            "titleSlug": "maximum-twin-sum-of-a-linked-list",
            "content": "<p>In a linked list of size <code>n</code>, where <code>n</code> is <strong>even</strong>, the <code>i<sup>th</sup></code> node (<strong>0-indexed</strong>) of the linked list is known as the <strong>twin</strong> of the <code>(n-1-i)<sup>th</sup></code> node, if <code>0 &lt;= i &lt;= (n / 2) - 1</code>.</p>\n\n<ul>\n\t<li>For example, if <code>n = 4</code>, then node <code>0</code> is the twin of node <code>3</code>, and node <code>1</code> is the twin of node <code>2</code>. These are the only nodes with twins for <code>n = 4</code>.</li>\n</ul>\n\n<p>The <strong>twin sum </strong>is defined as the sum of a node and its twin.</p>\n\n<p>Given the <code>head</code> of a linked list with even length, return <em>the <strong>maximum twin sum</strong> of the 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/2021/12/03/eg1drawio.png\" style=\"width: 250px; height: 70px;\" />\n<pre>\n<strong>Input:</strong> head = [5,4,2,1]\n<strong>Output:</strong> 6\n<strong>Explanation:</strong>\nNodes 0 and 1 are the twins of nodes 3 and 2, respectively. All have twin sum = 6.\nThere are no other nodes with twins in the linked list.\nThus, the maximum twin sum of the linked list is 6. \n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/12/03/eg2drawio.png\" style=\"width: 250px; height: 70px;\" />\n<pre>\n<strong>Input:</strong> head = [4,2,2,3]\n<strong>Output:</strong> 7\n<strong>Explanation:</strong>\nThe nodes with twins present in this linked list are:\n- Node 0 is the twin of node 3 having a twin sum of 4 + 3 = 7.\n- Node 1 is the twin of node 2 having a twin sum of 2 + 2 = 4.\nThus, the maximum twin sum of the linked list is max(7, 4) = 7. \n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/12/03/eg3drawio.png\" style=\"width: 200px; height: 88px;\" />\n<pre>\n<strong>Input:</strong> head = [1,100000]\n<strong>Output:</strong> 100001\n<strong>Explanation:</strong>\nThere is only one node with a twin in the linked list having twin sum of 1 + 100000 = 100001.\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 an <strong>even</strong> integer in the range <code>[2, 10<sup>5</sup>]</code>.</li>\n\t<li><code>1 &lt;= Node.val &lt;= 10<sup>5</sup></code></li>\n</ul>\n",
            "translatedTitle": "链表最大孪生和",
            "translatedContent": "<p>在一个大小为&nbsp;<code>n</code>&nbsp;且 <code>n</code>&nbsp;为&nbsp;<strong>偶数</strong> 的链表中，对于&nbsp;<code>0 &lt;= i &lt;= (n / 2) - 1</code>&nbsp;的 <code>i</code>&nbsp;，第&nbsp;<code>i</code>&nbsp;个节点（下标从 <strong>0</strong>&nbsp;开始）的孪生节点为第&nbsp;<code>(n-1-i)</code>&nbsp;个节点 。</p>\n\n<ul>\n\t<li>比方说，<code>n = 4</code>&nbsp;那么节点&nbsp;<code>0</code>&nbsp;是节点 <code>3</code>&nbsp;的孪生节点，节点 <code>1</code>&nbsp;是节点 <code>2</code>&nbsp;的孪生节点。这是长度为 <code><span style=\"\">n = 4</span></code>&nbsp;的链表中所有的孪生节点。</li>\n</ul>\n\n<p><strong>孪生和</strong>&nbsp;定义为一个节点和它孪生节点两者值之和。</p>\n\n<p>给你一个长度为偶数的链表的头节点&nbsp;<code>head</code>&nbsp;，请你返回链表的 <strong>最大孪生和</strong>&nbsp;。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例&nbsp;1：</strong></p>\n\n<p><img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/12/03/eg1drawio.png\" style=\"width: 250px; height: 70px;\"></p>\n\n<pre><b>输入：</b>head = [5,4,2,1]\n<b>输出：</b>6\n<strong>解释：</strong>\n节点 0 和节点 1 分别是节点 3 和 2 的孪生节点。孪生和都为 6 。\n链表中没有其他孪生节点。\n所以，链表的最大孪生和是 6 。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<p><img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/12/03/eg2drawio.png\" style=\"width: 250px; height: 70px;\"></p>\n\n<pre><b>输入：</b>head = [4,2,2,3]\n<b>输出：</b>7\n<strong>解释：</strong>\n链表中的孪生节点为：\n- 节点 0 是节点 3 的孪生节点，孪生和为 4 + 3 = 7 。\n- 节点 1 是节点 2 的孪生节点，孪生和为 2 + 2 = 4 。\n所以，最大孪生和为 max(7, 4) = 7 。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<p><img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/12/03/eg3drawio.png\" style=\"width: 200px; height: 88px;\"></p>\n\n<pre><b>输入：</b>head = [1,100000]\n<b>输出：</b>100001\n<strong>解释：</strong>\n链表中只有一对孪生节点，孪生和为 1 + 100000 = 100001 。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li>链表的节点数目是&nbsp;<code>[2, 10<sup>5</sup>]</code>&nbsp;中的&nbsp;<strong>偶数</strong>&nbsp;。</li>\n\t<li><code>1 &lt;= Node.val &lt;= 10<sup>5</sup></code></li>\n</ul>\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 pairSum(head: ListNode | null): number {\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 pair_sum(head: Option<Box<ListNode>>) -> i32 {\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 (pair-sum head)\n  (-> (or/c list-node? #f) exact-integer?)\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 pair_sum(Head :: #list_node{} | null) -> integer().\npair_sum(Head) ->\n  .",
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                "How can \"reversing\" a part of the linked list help find the answer?",
                "We know that the nodes of the first half are twins of nodes in the second half, so try dividing the linked list in half and reverse the second half.",
                "How can two pointers be used to find every twin sum optimally?",
                "Use two different pointers pointing to the first nodes of the two halves of the linked list. The second pointer will point to the first node of the reversed half, which is the (n-1-i)th node in the original linked list. By moving both pointers forward at the same time, we find all twin sums."
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