LeetCode每日一题(2130. Maximum Twin Sum of a Linked List)

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

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这题用递归的方式会报栈的深度超限， 用递归的好处是我们仅需要遍历一次。 既然递归不让用， 我们就用最直观的方法，先遍历链表生成数组， 然后计算每个 twin sum 找最大值

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impl Solution {
    pub fn pair_sum(mut head: Option<Box<ListNode>>) -> i32 {
        let mut sums = Vec::new();
        while let Some(node) = head {
            sums.push(node.val);
            head = node.next;
        }
        let mut ans = 0;
        for i in 0..sums.len() / 2 {
            ans = ans.max(sums[i] + sums[sums.len() - i - 1]);
        }
        ans
    }
}