LeetCode 328. Odd Even Linked List

一开始理解错题意，以为是根据节点内的值是奇偶来分成不同group，实际上是节点所在位置的奇偶（位置从1开始算起）。

 

思路正确后仍然RE的代码：

 1 class Solution {
 2 public:
 3     ListNode* oddEvenList(ListNode* head) {
 4         if(head == NULL) return head;
 5         if(head->next == NULL) return head;
 6         
 7         ListNode* odd = head, * even = head->next;
 8         odd->next = NULL;
 9         even->next = NULL;
10         ListNode* o_head = odd, * e_head = even;
11         while(odd && even && even->next){
12             odd->next = even->next;
13             odd = odd->next;
14             
15             even->next = odd->next;
16             even = even->next;
17         }
18         even->next = NULL;
19         odd->next = e_head->next;
20         return o_head->next;
21     }
22 };

起因是8、9行，因为odd和even本身已经是head和head->next两个节点，再设置next为NULL就相当于截断了原链表，even->next=NULL，进不了while循环。如果把18、19注释掉，不会RE。

 

 1 class Solution {
 2 public:
 3     ListNode* oddEvenList(ListNode* head) {
 4         if(head == NULL) return head;
 5         if(head->next == NULL) return head;
 6         
 7         ListNode* odd = head, * even = head->next;
 8         ListNode* o_head = odd, * e_head = even;
 9         
10         while(odd && even && even->next){
11             odd->next = even->next;
12             odd = odd->next;
13             
14             even->next = odd->next;
15             even = even->next;
16         }
17         // even->next = NULL;可有可无
18         odd->next = e_head;
19         return o_head;
20     }
21 };

 

posted @ 2016-03-14 19:20  co0oder  阅读(181)  评论(0)    收藏  举报