2130. Maximum Twin Sum of a Linked List | LeetCode Solution
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
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution
{
public:
ListNode *reverseLinkedList(ListNode *head)
{
if (head == NULL)
return head;
ListNode *pre = NULL;
ListNode *curr = head;
ListNode *next = NULL;
while (curr != NULL)
{
next = curr->next;
curr->next = pre;
pre = curr;
curr = next;
}
return pre;
}
int pairSum(ListNode *head)
{
stack<int> st;
int ans = INT_MIN;
ListNode *fast = head, *slow = head;
while (fast->next->next != NULL)
{
st.push(slow->val);
slow = slow->next;
fast = fast->next->next;
}
st.push(slow->val);
slow->next = reverseLinkedList(slow->next);
slow = slow->next;
ListNode *curr = head;
while (slow != NULL)
{
ans = max(curr->val + slow->val, ans);
slow = slow->next;
curr = curr->next;
}
return ans;
}
};
0 Comments
If you have any doubts/suggestion/any query or want to improve this article, you can comment down below and let me know. Will reply to you soon.