classSolution:defoddEvenList(self,head:Optional[ListNode])->Optional[ListNode]:ifnothead:returnNoneodd=head# 链表的第一个节点,即奇数位的第一个节点even=head.next# 链接表的第二个节点,即偶数位的第一个节点。如果链表只有两个元素,则不用进入下面的核心 while 循环even_head=even# 保存偶数位的第一个...
# 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]: # 初始化奇偶链表的哨兵结点,方便处理链表为空的情况 odd_dummy...
*/ class Solution { public: ListNode* oddEvenList(ListNode* head) {if(head == NULL || head -> next == NULL || head -> next -> next == NULL) return head; ListNode *tmp = head; int len = 0; while(tmp){tmp = tmp -> next; ++len; ...
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) space complexity and O(nodes) time complexity....
The first node is considered odd, the second node even and so on ... 解题思路:# 这道题很简单,迭代链表,将该链表奇数位节点和偶数位节点分别取出分隔成两个链表,然后将奇偶两个链表连接起来组成新链表,返回头节点即可。 需要记录偶数位节点的第一个节点,因为这是偶数链表的头节点,最后拼接链表时要用奇数...
The first node is considered odd, the second node even and so on … 解题思路: 这道题很简单,迭代链表,将该链表奇数位节点和偶数位节点分别取出分隔成两个链表,然后将奇偶两个链表连接起来组成新链表,返回头节点即可。 需要记录偶数位节点的第一个节点,因为这是偶数链表的头节点,最后拼接链表时要用奇数链表...
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) ...
每天一算:Odd Even Linked List LeetCode上第328号问题:Odd Even Linked List 题目 给定一个单链表,把所有的奇数节点和偶数节点分别排在一起。请注意,这里的奇数节点和偶数节点指的是节点编号的奇偶性,而不是节点的值的奇偶性。 请尝试使用原地算法完成。你的算法的空间复杂度应为 O(1),时间复杂度应为 O(...
将偶数号节点形成一条链,头结点为 evenHead。 将最后的奇数号节点 odd 与偶数号头节点 evenHead 建立连接。 js代码如下: varoddEvenList=function(head){if(head){// 奇数号链最后一个节点letodd=head// 偶数号链头letevenHead=head.next// 偶数号链最后一个节点leteven=evenHeadwhile(even){// 奇数号进...
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; ...