Re: [閒聊] 每日LeetCode已回收

看板Marginalman作者Rushia (みけねこ的鼻屎)時間2年前 (2023/05/18 00:18)推噓2(2推 0噓 0→)

留言2則, 2人參與討論串319/719 (看更多)

https://leetcode.com/problems/maximum-twin-sum-of-a-linked-list/description/ 2130. Maximum Twin Sum of a Linked List 給你一個偶數數量的鏈結串列，返回最大的一對串列和，一對串列為串列對半切之後從中心點分別往左和右有相同距離的Node。 Example 1: https://assets.leetcode.com/uploads/2021/12/03/eg1drawio.png

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: https://assets.leetcode.com/uploads/2021/12/03/eg3drawio.png

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. 思路： 1.一個指標每次走一步，一個每次走兩步，當快指標走完的時候慢指標會在中間，把慢指標停留的值紀錄在一個List。 2.慢指標繼續一步一步往後走並和 List 的尾端元素開始往前相加，並更新最大值。 Java Code： --------------------------------------------- class Solution { public int pairSum(ListNode head) { int max = 0; List<Integer> list = new ArrayList<>(); ListNode slow = head; ListNode fast = slow; while (fast != null) { list.add(slow.val); slow = slow.next; fast = fast.next.next; } int i = 1; while (slow != null) { max = Math.max(max, list.get(list.size() - i++) + slow.val); slow = slow.next; } return max; } } --------------------------------------------- 不想工作 -- https://i.imgur.com/tdaniED.jpg