2130. 链表最大孪生和

中等

题目描述

解题思路：
使用 快慢指针
慢指针和快指针初始都指向head
慢指针将数逐个加入到数组中，快指针步长为 2 right = right.next.next
当 right 为 null 时，慢指针指向第 size/2 + 1 个数

class Solution {
    public int pairSum(ListNode head) {
        ListNode left = head;
        ListNode right = head;
        int res = 0;
        ArrayList<Integer> list = new ArrayList<>();
        while (right != null){
            list.add(left.val);
            left = left.next;
            right = right.next.next;
        }
        int i = list.size() - 1;
        while (left != null){
            res = Math.max(list.get(i--) + left.val,res);
            left = left.next;
        }
        return res;
    }
}

function pairSum(head: ListNode | null): number {
    let left:ListNode | null = head;
    let right:ListNode | null = head;
    let res:number = 0;
    let list:number[] = [];
    while (right != null){
        list.push(left.val);
        left = left.next;
        right = right.next.next;
    }
    let i:number = list.length - 1;
    while (left != null){
        res = Math.max(list[i--] + left.val,res);
        left = left.next;
    }
    return res;
};