Reverse Linked List II

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.

思路：控制好链表的哨兵节点和尾部节点，要控制好使用到的每个节点的位置

java代码：

1. public ListNode reverseBetween(ListNode head, int m, int n) {
2. if(head==null || head.next==null) return head;
3. if(m==n) return head;
4. ListNode dummy = new ListNode(-1);
5. dummy.next = head;
6. ListNode pre = dummy;
7. ListNode p = head;
8. int i=1;
9. for(i=1;i<m;i++) {
10. pre = p;
11. p = p.next;
12. }
13. ListNode start = p;
14. while(i<n) {
15. p = p.next;  //p进入时为即将遍历到的上一个节点
16. start.next = p.next;
17. p.next = pre.next;
18. pre.next = p;
19. p = start;
20. i++;
21. }
22. return dummy.next;
23. }

C++代码：

1. ListNode *reverseBetween(ListNode *head, int m, int n) {
2. if(head==NULL || head->next==NULL) return head;
3. if(m==n) return head;
4. ListNode *dummy = new ListNode(0);
5. dummy->next=head;
6. ListNode *p=dummy;
7. int i=0;
8. for(i=1;i<m;i++) p=p->next;
9. head=p->next; //head为每次要遍历到的节点
10. i=i+1;
11. while(i<=n) {
12. ListNode *q=head->next;  
13. head->next=q->next;
14. q->next=p->next;
15. p->next=q;
16. i++;
17. }
18. return dummy->next;
19. }