2130.Maximum Twin Sum of a Linked List

tags: Medium,Linked List,Two Pointers,Stack

2130. Maximum Twin Sum of a Linked List

題目描述

In a linked list of size n, where n is even, the i^th 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.

範例

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

解答

Python

先貼個基本寫法

class Solution:
    def pairSum(self, head: Optional[ListNode]) -> int:
        arr = []
        
        while head:
            arr.append(head.val)
            head = head.next

        i, j = 0, len(arr) - 1
        mx = 0
        while i < j:
            mx = max(mx, arr[i] + arr[j])
            i += 1
            j -= 1

        return mx

class Solution:
    def pairSum(self, head: Optional[ListNode]) -> int:
        prev, slow, fast = None, head, head

        while fast and fast.next:
            fast = fast.next.next
            temp = slow.next
            slow.next = prev
            prev = slow
            slow = temp

        res = 0
        while slow:
            res = max(res, prev.val + slow.val)
            prev = prev.next
            slow = slow.next
            
        return res

Fast-slow Pointers
Time complexity:

$O(N)$
Space complexity:

$O(1)$

Ron ChenWed, May 17, 2023

Reference

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