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.
n = 4, then node 0 is the twin of node 3, and node 1 is the twin of node 2. These are the only nodes with twins for n = 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:
[2, 105].1 <= Node.val <= 105
class Solution {
public:
int pairSum(ListNode* head) {
if(!head)return 0;
// either reverse the linkedList and calculate or
// save it in a vector and calculate
vector<int> v;
while(head)
{
v.push_back(head->val);
head=head->next;
}
if(v.size()==1)return v[0];
int i=0,j=v.size()-1;
int maxSum=INT_MIN;
while(i<j)
{
maxSum=max(maxSum, v[i]+v[j]);
i++;
j--;
}
return maxSum;
}
};
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