leetcode_rotate_list

发表于

难度：Medium

解题思路：翻转很简单，把边界都考虑周到比较难。k＝0的情况如何处理，还有要对k取余。

dummy -> 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> NULL k=4
              ^    ^                   ^
              |    |                   |
             pre  first               second


  |---------------<--------------------|
  |                                    |
dummy -> 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7   NULL
              ^    ^                   ^
              |    |                   |
             pre  first               second

  |---------------<--------------------|
  |                                    |
dummy    1 -> 2 -> 3 ->  4 -> 5 -> 6 -> 7   NULL
  |           ^    ^                    ^
  |           |    |                    |
  |          pre  first               second
  |                |
  |-------->-------|       


         |-----------<--------------------------|
         |                                      |
dummy    1 -> 2 -> NULL    3 ->  4 -> 5 -> 6 -> 7   NULL
  |           ^            ^                    ^
  |           |            |                    |
  |          pre         first               second
  |                        |
  |-------->---------------|

代码如下：

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* rotateRight(ListNode* head, int k) {
        if(k == 0 || head == NULL ||head->next == NULL)  return head;
        
        ListNode *dummy = new ListNode(-1);
        dummy->next = head;
        
        int length = 0;
        ListNode *cur= dummy;
        while(cur ->next != NULL)
        {
            length++;
            cur = cur->next;
        }
        k = k % length;
        if(k == 0)  return head;
        
        ListNode *pre_first = dummy, *first = dummy, *second = dummy;
        for(int i = 0; i < k; i++)
        {
            if(second->next != NULL)
                second = second->next;
        }
        first = dummy->next;
        while(second->next != NULL)
        {
            pre_first = first;
            first = first->next;
            second = second->next;
        }
        if(pre_first == dummy)
            return dummy->next;
        second->next = dummy->next;
        dummy->next = first;
        pre_first->next = NULL;
        return dummy->next;
    }
};

代码结果：9ms，超过24.72%