LeetCode 25. Reverse Nodes in k-Group 

 1 class Solution {
 2 public:
 3     ListNode* reverseKGroup(ListNode* head, int k) {
 4         if(head == NULL || head->next == NULL || k == 1) return head;
 5         bool flag = true;
 6         ListNode* res = head;
 7         ListNode* prev = NULL;
 8         while(head){
 9             int num = k;
10             ListNode* start = head, * end = new ListNode(0);
11 
12             while(num != 0 && head){
13                 head = head->next;
14                 --num;
15                 if(num == 1) end = head;
16             }
17             
18             if(num != 0) break;//!!
19             end->next = NULL;//!!
20             
21             ListNode* p = head, * cur = start;
22             while(cur){
23                 ListNode* tmp = cur->next;
24                 cur->next = p;
25                 p = cur;
26                 cur = tmp;
27             }
28             if(prev) {prev->next = end;prev = start;}
29             
30             if(flag){
31                 res = p;//res = end;
32                 prev = start;
33                 flag = false;
34             }
35         }
36         
37         return res;
38     }
39 };

注意标记了!!的两行代码不能调换，否则RE.因为如果输入是[1,2], k = 3，那么在while循环里num=1时，end=NULL，end->next引起RE.

 

思路：先通过while循环得到要翻转的起始和末尾节点start、end，进行翻转，并和上一段节点的末尾prev，下一段节点的起始head相连。

 

代码又臭又长，应改进为更简洁的代码，留坑。

posted @ 2016-03-14 22:25  co0oder  阅读(130)  评论(0)  编辑  收藏  举报