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.

first 是第 m -1 个元素。 elementM是第m个元素。 后面将 cur -> second 变成 cur <- second。 最后 cur是第n个元素，second是第n + 1个元素。

 1 /**

 2  * Definition for singly-linked list.

 3  * public class ListNode {

 4  *     int val;

 5  *     ListNode next;

 6  *     ListNode(int x) {

 7  *         val = x;

 8  *         next = null;

 9  *     }

10  * }

11  */

12 public class Solution {

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

14         // Start typing your Java solution below

15         // DO NOT write main() function

16         ListNode header = new ListNode(-1);

17         header.next = head;

18         ListNode cur = header;

19         ListNode first = null;

20         ListNode elementM = null;

21         ListNode second = null;

22         int i = 0;

23         while(i < m){

24             first = cur;

25             cur = cur.next;

26             i ++;

27         }

28         elementM = cur;

29         second = cur.next;

30         while(i < n){

31             ListNode tmp = second.next;

32             second.next = cur;

33             cur = second;

34             second = tmp;

35             i ++;

36         }

37         first.next = cur;

38         if(elementM != null && elementM != second) elementM.next = second;

39         return header.next;

40     }

41 }

 

第三遍：

 1 public class Solution {

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

 3         ListNode header = new ListNode(-1);

 4         header.next = head;

 5         ListNode first = header, second = header;

 6         for(int i = 1; i < m; i ++){

 7             first = first.next; 

 8         }

 9         ListNode cur = first.next;

10         for(int i = 0; i < n; i ++){

11             second = second.next;

12         }

13         ListNode end = second;

14         second = second.next;

15         end.next = null;

16         while(cur != null){

17             ListNode tmp = cur.next;

18             cur.next = second;

19             second = cur;

20             cur = tmp;

21         }

22         first.next = end;

23         return header.next;

24     }

25 }