LeetCode Rotate List 

class Solution {
public:
    ListNode *rotateRight(ListNode *head, int k) {
        if (head == NULL) return NULL;
        ListNode *p = head, *q = head;

        int dst = k;
        while (dst--) {
            q = q->next;
            if (q == NULL) {
                q = head;
                k = k % (k - dst);
                dst = k;
            }
        }
        if (p == q) return head;
        while (q->next != NULL) {
            q = q->next;
            p = p->next;
        }
        ListNode *new_head = p->next;
        p->next = NULL;
        q->next = head;
        return new_head;
    }
};

中间进行适当的判断防止k>>list.length，而list.length又很大的情况下，重复遍历链表时间的浪费，这样在k>=list.length的情况下都只需完整遍历一次链表即可

第二轮：

Given a list, rotate the list to the right by k places, where k is non-negative.

For example:
Given 1->2->3->4->5->NULL and k = 2,
return 4->5->1->2->3->NULL.

感觉脑子进水了，链表可以直接移动啊，艹！

 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  // 10:00
10 class Solution {
11 public:
12     ListNode *rotateRight(ListNode *head, int k) {
13         if (head == NULL) return NULL;
14         
15         int cnt = 0;
16         ListNode* cur = head;
17         while (cur) {
18             cnt++;
19             cur = cur->next;
20         }
21         k = k % cnt;
22         
23         if (k == 0) return head;
24         
25         ListNode* rhead = reverse(head, cnt);
26         
27         int sk = k;
28         cur = rhead;
29         while (sk--) {
30             cur = cur->next;
31         }
32         
33         ListNode* rhead_part1 = reverse(rhead, k);
34         ListNode* rhead_part2 = reverse(cur, cnt - k);
35         
36         if (rhead_part1 != NULL) {
37             rhead->next = rhead_part2;
38         }
39         
40         return rhead_part1 == NULL ? rhead_part2 : rhead_part1;
41     }
42     
43     ListNode* reverse(ListNode* head, int k) {
44         ListNode* pre = NULL;
45         ListNode* cur = head;
46         while (k && cur) {
47             ListNode* next = cur->next;
48             cur->next = pre;
49             pre = cur;
50             cur = next;
51             k--;
52         }
53         return pre;
54     }
55 };

 赶紧再写一发：

 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         if (head == NULL) {
13             return NULL;
14         }
15         int cnt = 0;
16         ListNode* pre = NULL;
17         ListNode* cur = head;
18         while (cur) {
19             cnt++;
20             pre = cur;
21             cur = cur->next;
22         }
23         k = k % cnt;
24         if (k == 0) {
25             return head;
26         }
27         
28         ListNode* last = pre;
29         pre = NULL;
30         cur = head;
31         int rk = cnt - k;
32         while (rk--) {
33             pre = cur;
34             cur = cur->next;
35         }
36         last->next= head;
37         pre->next = NULL;
38         return cur;
39     }
40 };

 

posted @ 2014-04-17 16:55  卖程序的小歪  阅读(158)  评论(0)  编辑  收藏  举报