Given a singly linked list L: L0→L1→…→Ln-1→Ln, reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→… You may not modify the values in the list's nodes, only nodes itself may be changed. 题意 给定一个单链表 L:L0→L1→…→L
# Definition for singly-linked list. # class ListNode: # def __init__(self, val=0, next=None): # self.val = val # self.next = next class Solution: def reverseBetween(self, head: ListNode, left: int, right: int) -> ListNode: def reverse_linked_list(head: ListNode): # 也可以使...
/*** Definition for singly-linked list.* struct ListNode {* int val;* ListNode *next;* ListNode(int x) : val(x), next(NULL) {}* };*/classSolution{public:ListNode*deleteDuplicates(ListNode* head){if(head ==NULL)returnNULL;ListNodedummy(0);dummy.next = head;ListNode *node = &dummy;...
Given a singly linked list, determine if it is a palindrome. Example 1: Input:1->2Output:false Example 2: Input:1->2->2->1Output:true Follow up: Could you do it in O(n) time and O(1) space? 一开始不理解回文是什么意思,于是百度了一波。总结一下,回文就是对称的意思。 由于链表的...
# Definition for singly-linked list. # class ListNode: # def __init__(self, val=0, next=None): # self.val = val # self.next = next class Solution: def deleteDuplicates(self, head: Optional[ListNode]) -> Optional[ListNode]: # 如果是空链表,则直接返回 None if head is None: return...
https://leetcode.cn/problems/intersection-of-two-linked-lists/description/ 代码语言:javascript 代码运行次数:0 运行 AI代码解释 /** * Definition for singly-linked list. * function ListNode(val) { * this.val = val; * this.next = null; ...
* Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode* reverseList(ListNode* head) { if(head == NULL) ...
* Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */classSolution{publicbooleanisPalindrome(ListNode head){int sum1=0;if(head==null||head.next==null)returntrue;int count=1;ListNode temp=head;while(temp!
* Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */classSolution{publicListNodemergeKLists(ListNode[] lists){intlength=lists.length;if(length==0)returnnull;if(length==1)returnlists[0];ListNoderesult=lists...
// Definition for singly-linked list.#[derive(PartialEq, Eq, Clone, Debug)]pub struct ListNode { pub val: i32, pub next: Option<Box<ListNode>>} 1. 需要理解的是next字段的类型为Option<Box<ListNode>>,这个类型不存在任何的引用,暗含的意思就是:链表头是整个链表的拥有者,负责整个链表所占据内存的...