92. Reverse Linked List II【Medium】

Reverse a linked list from position m to n. Do it in-place and in one-pass.

For example:
Given 1->2->3->4->5->NULL, m = 2 and n = 4,

return 1->4->3->2->5->NULL.

Note:
Given m, n satisfy the following condition:
1 ≤ m ≤ n ≤ length of list.

解法一：

 1 class Solution {
 2 public:
 3     ListNode* reverseBetween(ListNode* head, int m, int n) {
 4         if (head == NULL || head->next == NULL) {
 5             return head;
 6         }
 7         
 8         ListNode * dummy = new ListNode(INT_MIN);
 9         dummy->next = head;
 
10         ListNode * mth_prev = findkth(dummy, m - 1);
11         ListNode * mth = mth_prev->next;
12         ListNode * nth = findkth(dummy, n);
13         ListNode * nth_next = nth->next;
14         nth->next = NULL;
15         
16         reverseList(mth);
17         
18         mth_prev->next = nth;
 
19         mth->next = nth_next;
20         
21         return dummy->next;
22     }
23     
24     ListNode *findkth(ListNode *head, int k) 
25     {
26         for (int i = 0; i < k; i++) {
27             if (head == NULL) {
28                 return NULL;
29             }
30             head = head->next;
31         }
32         return head;
33     }
34     
35     ListNode * reverseList(ListNode * head)
36     {
37         ListNode * pre = NULL;
38         while (head != NULL) {
39             ListNode * next = head->next;
40             head->next = pre;
41             pre = head;
42             head = next;
43         }
44         
45         return pre;
46     }
47 };

解法二：

 1 class Solution {
 2 public:
 3     ListNode* reverseBetween(ListNode* head, int m, int n) {
 4         if (head == NULL || head->next == NULL) {
 5             return head;
 6         }
 7         
 8         ListNode * dummy = new ListNode(INT_MIN);
 9         dummy->next = head;
10         head = dummy;
11         
12         for (int i = 1; i < m; ++i) {
13             if (head == NULL) {
14                 return NULL;
15             }
16             head = head->next;
17         }
18         
19         ListNode * premNode = head;
20         ListNode * mNode = head->next;
21         ListNode * nNode = mNode;
22         ListNode * postnNode = mNode->next;
23         
24         for (int i = m; i < n; ++i) {
25             if (postnNode == NULL) {
26                 return NULL;
27             }
28             ListNode * temp = postnNode->next;
29             postnNode->next = nNode;
30             nNode = postnNode;
31             postnNode = temp;
32         }
33         
34         mNode->next = postnNode;
35         premNode->next = nNode;
36         
37         return dummy->next;
38     }
39 };

解法三：

 1 public ListNode reverseBetween(ListNode head, int m, int n) {
 2     if(head == null) return null;
 3     ListNode dummy = new ListNode(0); // create a dummy node to mark the head of this list
 4     dummy.next = head;
 5     ListNode pre = dummy; // make a pointer pre as a marker for the node before reversing
 6     for(int i = 0; i<m-1; i++) pre = pre.next;
 7     
 8     ListNode start = pre.next; // a pointer to the beginning of a sub-list that will be reversed
 9     ListNode then = start.next; // a pointer to a node that will be reversed
10     
11     // 1 - 2 -3 - 4 - 5 ; m=2; n =4 ---> pre = 1, start = 2, then = 3
12     // dummy-> 1 -> 2 -> 3 -> 4 -> 5
13     
14     for(int i=0; i<n-m; i++)
15     {
16         start.next = then.next;
17         then.next = pre.next;
18         pre.next = then;
19         then = start.next;
20     }
21     
22     // first reversing : dummy->1 - 3 - 2 - 4 - 5; pre = 1, start = 2, then = 4
23     // second reversing: dummy->1 - 4 - 3 - 2 - 5; pre = 1, start = 2, then = 5 (finish)
24     
25     return dummy.next;
26     
27 }

參考了@ardyadipta 的代碼，Simply just reverse the list along the way using 4 pointers: dummy, pre, start, then

92. Reverse Linked List II【Medium】