* public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ public class Solution { public ListNode oddEvenList(ListNode head) { if (head == null || head.next == null) return head; ListNode even = head.next; ListNode evenNext = even; ListNo...
even 初始化为链表的第二个节点,即第一个偶数节点。 even_head 保存偶数节点链表的头,以便后续将其连接到奇数节点链表的尾部 迭代遍历链表: 使用while even and even.next 遍历链表,确保每次操作都将奇数节点连接到下一个奇数节点,偶数节点连接到下一个偶数节点。 odd.next = even.next:将当前奇数节点的下一个...
Copy classSolution{publicListNodeoddEvenList(ListNode head){if(head ==null|| head.next ==null|| head.next.next ==null)returnhead;//如果该链表内节点数在两个及以下直接返回头节点ListNodetmp=head.next;//暂存偶节点的第一个ListNodeodd=head;//奇节点的第一个ListNodeeven=head.next;//偶节点的第一...
https://leetcode.com/problems/odd-even-linked-list/ 建两个dummy node,一个维护奇数节点构成的链表,另一个维护偶数节点构成的链表,最后再拼接起来。注意偶数节点构成的链表最后一个节点的next域要置空,否则可能会导致出现环路。 classSolution{public:ListNode*oddEvenList(ListNode* head){if(!head || !head-...
LeetCode-Odd Even Linked List Description: Given a singly linked list, group all odd nodes together followed by the even nodes. Please note here we are talking about the node number and not the value in the nodes. You should try to do it in place. The program should run in O(1) ...
classSolution {public: ListNode* oddEvenList(ListNode*head) {if(!head || !head->next)returnhead; ListNode*odd = head, *even = head->next, *even_head =even;while(even && even->next) { odd= odd->next = even->next; even= even->next = odd->next; ...
https://leetcode.com/problems... # Definition for singly-linked list. # class ListNode: # def __init__(self, x): # self.val = x # self.next = None class Solution: def oddEvenList(self, head: ListNode) -> ListNode: # If zero, one or two elements, then solved ...
两年前的代码. 参考 https://leetcode.com/problems/odd-even-linked-list/solution/ classSolution{ public: ListNode*oddEvenList(ListNode*head) { if(!head||!head->next) returnhead; ListNode*odd=head,*even=head->next,*evenHead=even; ...
*/structListNode*oddEvenList(structListNode*head){if((head==NULL)||(head->next==NULL))returnhead;structListNode*slow=head;structListNode*fast=head->next->next;//fastpre - is the just one node previous to faststructListNode*fastpre=head->next;while(slow&&fast){//copy current 'fast' in ...
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