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...
* 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...
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-...
1classSolution {2public:3ListNode* oddEvenList(ListNode*head) {4if(head == NULL)returnhead;5if(head->next == NULL)returnhead;67ListNode* odd = head, * even = head->next;8ListNode* o_head = odd, * e_head =even;910while(odd && even && even->next){11odd->next = even->next;...
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) ...
The first node is considered odd, the second node even and so on ... Credits: Special thanks to@aadarshjajodiafor adding this problem and creating all test cases. 这道题给了我们一个链表,让我们分开奇偶节点,所有奇节点在前,偶节点在后。我们可以使用两个指针来做,pre指向奇节点,cur指向偶节点,然...
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(...
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 ...
Can you solve this real interview question? Even Odd Tree - A binary tree is named Even-Odd if it meets the following conditions: * The root of the binary tree is at level index 0, its children are at level index 1, their children are at level index 2,
LeetCode: 975. Odd Even Jump 方国正 计算机硕士 来自专栏 · 进阶算法 题目描述:奇偶跳 给定一个整形数组A,从某一个下标开始,你可以进行一系列的跳跃。其中,第(1,3,5,...)次跳跃被称为奇数跳,而第(2,4,6,...)次跳跃被称为偶数跳。 你可以从下标 i 向前跳跃到下标 j ,其中 (i<j) 。本次...