LeetCode: Rotate List

两个边界条件没考虑到，少数次过

 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 *rotateRight(ListNode *head, int k) {
12         // Start typing your C/C++ solution below
13         // DO NOT write int main() function
14         ListNode *tmp = head;
15         if (!head) return NULL;
16         int n = 0;
17         while (tmp) {
18             n++;
19             tmp = tmp->next;
20         }
21         k %= n;
22         ListNode *p = head;
23         tmp = head;
24         for (int i = 0; i < n-k-1; i++) tmp = tmp->next;
25         ListNode *q = tmp->next;
26         if (!q) return head;
27         head = q;
28         tmp->next = NULL;
29         while (q->next) q = q->next;
30         q->next = p;
31         return head;
32     }
33 };

 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     public int val;
 5  *     public ListNode next;
 6  *     public ListNode(int x) { val = x; }
 7  * }
 8  */
 9 public class Solution {
10     public ListNode RotateRight(ListNode head, int k) {
11         ListNode tmp = head;
12         if (head == null) return null;
13         int n = 0;
14         while (tmp != null) {
15             n++;
16             tmp = tmp.next;
17         }
18         k %= n;
19         ListNode p = head;
20         tmp = head;
21         for (int i = 0; i < n-k-1; i++) tmp = tmp.next;
22         ListNode q = tmp.next;
23         if (q == null) return head;
24         head = q;
25         tmp.next = null;
26         while (q.next != null) q = q.next;
27         q.next = p;
28         return head;
29     }
30 }

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posted @ 2013-04-20 15:16 ying_vincent 阅读(137) 评论(0) 编辑收藏举报

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