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...
() even_dummy: ListNode = ListNode() # 初始化每个链表的尾部结点,方便使用尾插法插入结点 odd_tail, even_tail = odd_dummy, even_dummy # is_odd 表示当前结点的下标是否为奇数 is_odd: bool = True # 当还有结点时,继续分组到对应的链表 while head: if is_odd: # 如果下标为奇数,则将 head ...
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-...
(1)先递归oddEvenList(head -> next -> next),链表此时为2->1->(递归结果:3->6->7->5->4)(2)把2插入到3(递归结果奇数位的第一个位置)前面,把1插入到5前面(递归结果偶数位的第一个位置)此法缺点,每次递归都要求当前链表长度,时间复杂度肯定超O(n)了,不推荐...
func oddEvenList(head *ListNode) *ListNode { if head == nil { return head } evenHead := head.Next odd, even := head, evenHead for even != nil && even.Next != nil { odd.Next, odd = even.Next, even.Next even.Next, even = odd.Next, odd.Next ...
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(...
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...
# Link the last odd node to the head of the even nodes list odd.next = even_root # Return the reordered list starting from the head return head ``` ## References - [Leetcode Odd Even Linked List Problem](https://leetcode.com/problems/swap-nodes-in-pairs) - [파이썬 알고...
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...
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){// 奇数号进行链接odd.next=even.next// 记录最后一个奇数...