Re: [閒聊] 每日leetcode已回收

看板Marginalman作者sustainer123 (caster )時間1年前 (2024/05/06 12:08)推噓6(6推 0噓 8→)

留言14則, 6人參與討論串190/1548 (看更多)

※ 引述《Rushia (早瀬ユウカの体操服 )》之銘言： : https://leetcode.com/problems/remove-nodes-from-linked-list/description/ : 2487. Remove Nodes From Linked List : 給你一個鏈結串列，移除串列中所有右邊存在比他大的數字的的節點。 : https://assets.leetcode.com/uploads/2022/10/02/drawio.png

: 思路: : 1.題目其實是在求一個最長遞減串列，我們將原串列做反轉之後把遞增的串起來，再把 : 串起來的串列也反轉即可。 : py code: : ---------------------------------------------- : class Solution: : def removeNodes(self, head: Optional[ListNode]) -> Optional[ListNode]: : def reverse(node: Optional[ListNode]): : curr, prev = node, None : while curr: : tmp = curr.next : curr.next = prev : prev = curr : curr = tmp : return prev : head = reverse(head) : dummy = ListNode(-1) : curr = dummy : while head: : if head.val >= curr.val: : curr.next = head : curr = curr.next : head = head.next : curr.next = None : return reverse(dummy.next) : ---------------------------------------------- 照抄大老思路 Java Code: /** * 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 ListNode reverse(ListNode node){ ListNode pre = null; ListNode cur = node; while(cur != null){ ListNode tmp = cur.next; cur.next = pre; pre = cur; cur = tmp; } return pre; } public ListNode removeNodes(ListNode head) { head = reverse(head); ListNode dummy = new ListNode(); ListNode cur = dummy; while(head != null){ if(head.val >= cur.val){ cur.next = head; cur = head; } head = head.next; } cur.next = null; return reverse(dummy.next); } } 1 Java函式裡面不能放函式 2 分號 3 static可以理解成全域變數ㄇ -- ※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 114.136.198.113 (臺灣) ※ 文章網址: https://www.ptt.cc/bbs/Marginalman/M.1714968533.A.6DA.html

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