LeetCode题解：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.

题意：给定整数m,n和单链表头结点，将链表从m->n的结点顺序逆转，O（n）时间要求

解决思路：先找到m->n子链表头结点的前驱结点，然后改变子链表的顺序就可以了。如题中例子，找到2->3->4这个子链表，然后得到3->2->4，然后4->3->2

代码：

public class Solution {    public ListNode reverseBetween(ListNode head, int m, int n) {        if(head == null) return null;        ListNode dummy = new ListNode(0); // 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 = 0; i<m-1; 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=0; 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;    }}