2130. Maximum Twin Sum of a Linked List
In a linked list of size n, where n is even, the i t h i^{th} ith node (0-indexed) of the linked list is known as the twin of the ( n ? 1 ? i ) t h (n-1-i)^{th} (n?1?i)th node, if 0 <= i <= (n / 2) - 1.
- For example, if 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.
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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.文章來(lái)源：http://www.zghlxwxcb.cn/news/detail-799319.html
Constraints:
- The number of nodes in the list is an even integer in the range [ 2 , 1 0 5 ] [2, 10^5] [2,105].
- 1 < = N o d e . v a l < = 1 0 5 1 <= Node.val <= 10^5 1<=Node.val<=105
From: LeetCode
Link: 2130. Maximum Twin Sum of a Linked List文章來(lái)源地址http://www.zghlxwxcb.cn/news/detail-799319.html
Solution:
Ideas：
- Use the two-pointer technique to find the middle of the linked list.
- Reverse the second half of the list.
- Sum the values of the corresponding nodes from the start and reversed second half to find the maximum sum.
- Return the maximum sum found.
Code:
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
int pairSum(struct ListNode* head) {
if (head == NULL || head->next == NULL) return 0;
// Step 1: Find the middle of the linked list
struct ListNode *slow = head, *fast = head;
while (fast && fast->next) {
slow = slow->next;
fast = fast->next->next;
}
// Step 2: Reverse the second half of the linked list
struct ListNode *prev = NULL, *next = NULL;
while (slow) {
next = slow->next;
slow->next = prev;
prev = slow;
slow = next;
}
// Step 3: Pairwise sum the values from the start and end to find the maximum sum
int maxSum = 0;
struct ListNode *start = head, *end = prev;
while (end) {
maxSum = maxSum > (start->val + end->val) ? maxSum : (start->val + end->val);
start = start->next;
end = end->next;
}
// Step 4: Return the maximum sum
return maxSum;
}
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