2130. Maximum Twin Sum of a Linked List
Medium
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 node0
is the twin of node3
, and node1
is 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
把數值都存進陣列後暴力解。
/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func pairSum(head *ListNode) int {
arr := make([]int, 0)
for head != nil {
arr = append(arr, head.Val)
head = head.Next
}
ans := 0
for i:=0; i < len(arr) / 2 ; i++ {
if arr[i] + arr[len(arr) - 1 - i] > ans {
ans = arr[i] + arr[len(arr) - 1 - i]
}
}
return ans
}