LeetCode 2130. Maximum Twin Sum of a Linked List

原题链接在这里：https://leetcode.com/problems/maximum-twin-sum-of-a-linked-list/

题目：

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 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:

• The number of nodes in the list is an even integer in the range [2, 105].
• 1 <= Node.val <= 105

题解：

Find mid node, reverse the list after midNext, its new head is newHead.

Now we have head and newHead, they are twins.

Iterate through 2 half lists and update the maximum.

Time Complexity: O(n). n is the length of list. findMid takes O(n). reverse takes O(n).

Space: O(1).

AC Java:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public int pairSum(ListNode head) {
        if(head == null){
            return 0;
        }
        
        ListNode mid = findMid(head);
        ListNode midNext = mid.next;
        mid.next = null;
        midNext = reverse(midNext);
        int res = Integer.MIN_VALUE;
        while(head != null && midNext != null){
            res = Math.max(res, head.val + midNext.val);
            head = head.next;
            midNext = midNext.next;
        }
        
        return res;
    }
    
    private ListNode findMid(ListNode head){
        if(head == null || head.next == null){
            return head;
        }
        
        ListNode runner = head;
        ListNode walker = head;
        while(runner.next != null && runner.next.next != null){
            walker = walker.next;
            runner = runner.next.next;
        }
        
        return walker;
    }
    
    private ListNode reverse(ListNode head){
        if(head == null || head.next == null){
            return head;
        }
        
        ListNode tail = head;
        ListNode cur = head;
        ListNode pre;
        ListNode temp;
        while(tail.next != null){
            pre = cur;
            cur = tail.next;
            temp = cur.next;
            cur.next = pre;
            tail.next = temp;
        }
        
        return cur;
    }
}

类似Palindrome Linked List.