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：

 class Solution {

 public:

     ListNode* reverseBetween(ListNode* head, int m, int n) {

         if (head == NULL || head->next == NULL) {

             return head;

         }

         ListNode * dummy = new ListNode(INT_MIN);

         dummy->next = head;

         ListNode * mth_prev = findkth(dummy, m - );

         ListNode * mth = mth_prev->next;

         ListNode * nth = findkth(dummy, n);

         ListNode * nth_next = nth->next;

         nth->next = NULL;

         reverseList(mth);

         mth_prev->next = nth;

         mth->next = nth_next;

         return dummy->next;

     }

     ListNode *findkth(ListNode *head, int k)

     {

         for (int i = ; i < k; i++) {

             if (head == NULL) {

                 return NULL;

             }

             head = head->next;

         }

         return head;

     }

     ListNode * reverseList(ListNode * head)

     {

         ListNode * pre = NULL;

         while (head != NULL) {

             ListNode * next = head->next;

             head->next = pre;

             pre = head;

             head = next;

         }

         return pre;

     }

 };

解法二：

 class Solution {

 public:

     ListNode* reverseBetween(ListNode* head, int m, int n) {

         if (head == NULL || head->next == NULL) {

             return head;

         }

         ListNode * dummy = new ListNode(INT_MIN);

         dummy->next = head;

         head = dummy;

         for (int i = ; i < m; ++i) {

             if (head == NULL) {

                 return NULL;

             }

             head = head->next;

         }

         ListNode * premNode = head;

         ListNode * mNode = head->next;

         ListNode * nNode = mNode;

         ListNode * postnNode = mNode->next;

         for (int i = m; i < n; ++i) {

             if (postnNode == NULL) {

                 return NULL;

             }

             ListNode * temp = postnNode->next;

             postnNode->next = nNode;

             nNode = postnNode;

             postnNode = temp;

         }

         mNode->next = postnNode;

         premNode->next = nNode;

         return dummy->next;

     }

 };

解法三：

 public ListNode reverseBetween(ListNode head, int m, int n) {

     if(head == null) return null;

     ListNode dummy = new ListNode(); // create a dummy node to mark the head of this list

     dummy.next = head;

     ListNode pre = dummy; // make a pointer pre as a marker for the node before reversing

     for(int i = ; i<m-; i++) pre = pre.next;

     ListNode start = pre.next; // a pointer to the beginning of a sub-list that will be reversed

     ListNode then = start.next; // a pointer to a node that will be reversed

     // 1 - 2 -3 - 4 - 5 ; m=2; n =4 ---> pre = 1, start = 2, then = 3

     // dummy-> 1 -> 2 -> 3 -> 4 -> 5

     for(int i=; i<n-m; i++)

     {

         start.next = then.next;

         then.next = pre.next;

         pre.next = then;

         then = start.next;

     }

     // first reversing : dummy->1 - 3 - 2 - 4 - 5; pre = 1, start = 2, then = 4

     // second reversing: dummy->1 - 4 - 3 - 2 - 5; pre = 1, start = 2, then = 5 (finish)

     return dummy.next;

 }

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

巴特西

92. Reverse Linked List II【Medium】

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