328. Odd Even Linked List

Posted 2021-01-19 ming-1012

题目链接：https://leetcode.com/problems/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) space complexity and O(nodes) time complexity.

Example 1:

Input: 1->2->3->4->5->NULL
Output: 1->3->5->2->4->NULL

Example 2:

Input: 2->1->3->5->6->4->7->NULL
Output: 2->3->6->7->1->5->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 ...

 

思路：

• 本题思路较直接，即将链表中处于奇数位置（1、3、...）的结点和处于偶数位置（2、4、...）的结点进行分离，形成两条新的链表；
• 用指针oddHead指向由奇数结点组成的链表头结点；用指针evenHead指向由偶数结点组成的链表头结点；
• 奇数结点组成的链表尾结点的next域指向偶数结点组成的链表的头结点即为所求。

 

编码如下：

 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 class Solution {
10 public:
11     ListNode* oddEvenList(ListNode* head) {
12         if (head == nullptr) return nullptr;
13         
14         ListNode *oddHead = nullptr;
15         ListNode *pOdd = nullptr;
16         ListNode *evenHead = nullptr;
17         ListNode *pEven = nullptr;
18         
19         int i = 0;
20         ListNode *cur = head;
21         while (nullptr != cur)
22         {
23             i++;
24             ListNode *p = cur->next;
25             if ( i & 1 == 1)    // i为奇数时
26             {
27                 if (nullptr == oddHead)
28                 {
29                     oddHead = cur;
30                     pOdd = oddHead;
31                     pOdd->next = nullptr;
32                 }
33                 else
34                 {
35                     pOdd->next = cur;
36                     pOdd = pOdd->next;
37                     pOdd->next = nullptr;
38                 }
39             }
40             else
41             {
42                 if (nullptr == evenHead)
43                 {
44                     evenHead = cur;
45                     pEven = evenHead;
46                     pEven->next = nullptr;
47                 }
48                 else
49                 {
50                     pEven->next = cur;
51                     pEven = pEven->next;
52                     pEven->next = nullptr;
53                 }
54             }
55             
56             cur = p;
57         }
58         
59         pOdd->next = evenHead;
60         
61         return head;
62     }
63 };