Leetcode ☞ 328. Odd Even Linked List ☆

328. Odd Even Linked List

Question

Total Accepted: 18870 Total Submissions: 50234 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 ...

我的AC（4ms，最快一批）：

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     struct ListNode *next; * }; */struct ListNode* oddEvenList(struct ListNode* head) {    if(!head || !head->next)        return head;            struct ListNode *oddList, *evenList, *evenHead;    oddList = head;    evenList = evenHead = head->next;     while(oddList->next && evenList->next){        oddList->next = oddList->next->next;        evenList->next = evenList->next->next;        oddList = oddList->next;        evenList = evenList->next;    }    oddList->next = evenHead;        return head;}

分析： 

1、if(!head || !head->next)一定要加，其实if(!head)就够了，防[]

2、链表的循环里总忘记并没有“++”，需要自己加list = list->next;