LeetCode 328. Odd Even Linked List

Posted 2020-06-13

题目链接：

https://leetcode.com/problems/odd-even-linked-list/

 

328. Odd Even Linked List

My Submissions

Question

Total Accepted: 13612 Total Submissions: 35990 Difficulty: Easy

 

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.

 

Example:
Given 1->2->3->4->5->NULL,
return 1->3->5->2->4->NULL.

Note:
The relative order inside both the even and odd groups should remain as it was in the input. 
The first node is considered odd, the second node even and so on ...

Credits:
Special thanks to @aadarshjajodia for adding this problem and creating all test cases.

 

Subscribe to see which companies asked this question

 

做了一天，第一次调试链表相关的程序，好不适应。

 

 1 /**
 2  * Definition for singly-linked list.
 3  * struct ListNode {
 4  *     int val;
 5  *     ListNode *next;
 6  *     ListNode(int x) : val(x), next(NULL) {}
 7  * };
 8  */
 9  /***********************************
10   * 
11   * 思路：
12   * 1)记录下偶节点的头evenhead
13   * 2)扫一遍链表，根据节点的奇偶，形成 odd,even 两个链表
14   * 3)把偶链表的头evenhead 接到 奇链表后
15   * 
16   * *******************************************
17   */
18 class Solution {
19 public:
20     ListNode* oddEvenList(ListNode* head) {
21         if(head == NULL || head->next == NULL){
22             return head;
23         }
24         ListNode* odd = head;
25         ListNode* evenhead = head->next;
26         ListNode* even = evenhead;
27         //return evenhead;
28         ListNode *phead = head->next;
29         int count = 2;
30         while(phead->next != NULL){
31             if(count & 1){
32                 even->next = phead->next;
33                 even = even->next;
34             }
35             else{
36                 odd->next = phead->next;
37                 odd = odd->next;
38             }
39             phead = phead->next;
40             count++;
41         }
42         odd->next = evenhead;
43         even->next = NULL;
44         return head;
45     }
46 };