[LeetCode] 2130. Maximum Twin Sum of a Linked List
In a linked list of size n, where n is even, the ith node (0-indexed) of the linked list is known as the twin of the (n-1-i)th node, if 0 <= i <= (n / 2) - 1.
- For example, if
n = 4, then node0is the twin of node3, and node1is the twin of node2. These are the only nodes with twins forn = 4.
The twin sum is defined as the sum of a node and its twin.
Given the head of a linked list with even length, return the maximum twin sum of the linked list.
Example 1:
Input: head = [5,4,2,1] Output: 6 Explanation: Nodes 0 and 1 are the twins of nodes 3 and 2, respectively. All have twin sum = 6. There are no other nodes with twins in the linked list. Thus, the maximum twin sum of the linked list is 6.
Example 2:
Input: head = [4,2,2,3] Output: 7 Explanation: The nodes with twins present in this linked list are: - Node 0 is the twin of node 3 having a twin sum of 4 + 3 = 7. - Node 1 is the twin of node 2 having a twin sum of 2 + 2 = 4. Thus, the maximum twin sum of the linked list is max(7, 4) = 7.
Example 3:
Input: head = [1,100000] Output: 100001 Explanation: There is only one node with a twin in the linked list having twin sum of 1 + 100000 = 100001.
Constraints:
- The number of nodes in the list is an even integer in the range
[2, 105]. 1 <= Node.val <= 105
链表最大孪生和。
在一个大小为 n 且 n 为 偶数 的链表中，对于 0 <= i <= (n / 2) - 1 的 i ，第 i 个节点（下标从 0 开始）的孪生节点为第 (n-1-i) 个节点 。
比方说，n = 4 那么节点 0 是节点 3 的孪生节点，节点 1 是节点 2 的孪生节点。这是长度为 n = 4 的链表中所有的孪生节点。
孪生和 定义为一个节点和它孪生节点两者值之和。给你一个长度为偶数的链表的头节点 head ，请你返回链表的 最大孪生和 。
来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/maximum-twin-sum-of-a-linked-list
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这道题思路跟234题一样，思路还是找到链表的中点，反转后半部分，然后前半后半同时遍历，以找到最大孪生和。
时间O(n)
空间O(1)
Java实现
1 /** 2 * Definition for singly-linked list. 3 * public class ListNode { 4 * int val; 5 * ListNode next; 6 * ListNode() {} 7 * ListNode(int val) { this.val = val; } 8 * ListNode(int val, ListNode next) { this.val = val; this.next = next; } 9 * } 10 */ 11 class Solution { 12 public int pairSum(ListNode head) { 13 // corner case 14 if (head == null) { 15 return 0; 16 } 17 18 // normal case 19 ListNode slow = head; 20 ListNode fast = head; 21 while (fast.next != null && fast.next.next != null) { 22 slow = slow.next; 23 fast = fast.next.next; 24 } 25 ListNode middle = slow; 26 middle.next = reverse(middle.next); 27 ListNode p1 = head; 28 ListNode p2 = middle.next; 29 // 也许要把两部分链表分开，这里不分开也能AC 30 // middle.next = null; 31 int max = 0; 32 while (p1 != null && p2 != null) { 33 max = Math.max(max, p1.val + p2.val); 34 p1 = p1.next; 35 p2 = p2.next; 36 } 37 return max; 38 } 39 40 private ListNode reverse(ListNode head) { 41 ListNode pre = null; 42 while (head != null) { 43 ListNode next = head.next; 44 head.next = pre; 45 pre = head; 46 head = next; 47 } 48 return pre; 49 } 50 }
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