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→…→Ln-1→Ln , 将其重新排列后变为: L0→Ln→L1→Ln-...
/*** 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;...
Design your implementation of the linked list. You can choose to use the singly linked list or the doubly linked list. A node in a singly linked list should have two attributes:valandnext.valis the value of the current node, andnextis a pointer/reference to the next node. If you want ...
# 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. # class ListNode: # def __init__(self, val=0, next=None): # self.val = val # self.next = next class Solution: def oddEvenList(self, head: Optional[ListNode]) -> Optional[ListNode]: if not head or not head.next: return head odd,even=head,...
Given the head of a singly linked list, group all the nodes with odd indices together followed by the nodes with even indices, and return the reordered list. The first node is considered odd, and the second node is even, and so on. ...
* 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.#[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...
* 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...
Design your implementation of the linked list. You can choose to use a singly or doubly linked list.