Problem
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
Solution
/**
* Definition for singly-linked list.
* function ListNode(val, next) {
* this.val = (val===undefined ? 0 : val)
* this.next = (next===undefined ? null : next)
* }
*/
/**
* @param {ListNode} head
* @return {number}
*/
var pairSum = function(head) {
var num = 0;
var node = head;
while (node) {
num++;
node = node.next;
}
var stack = [];
var sum = 0;
var i = 0;
while (i < num) {
if (i < num / 2) {
stack.push(head.val);
} else {
sum = Math.max(sum, head.val + stack.pop());
}
i += 1;
head = head.next;
}
return sum;
};
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).