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->NULLm = 2 and n = 4,

return 1->4->3->2->5->NULL.

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

 

将链表m-n之间的节点反转。

 

而且已经规定了  1 ≤ m ≤ n ≤ length of list 。 所以也没什么难点。

当然代码还可以化简。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode reverseBetween(ListNode head, int m, int n) {
    if( head == null || head.next == null || m==n)
        return head;
    ListNode bef = head;
    for( int i = 2;i<m;i++)
        bef = bef.next;
    ListNode start = null;
    if( m < 2){
        bef = null;
        start = head;
    }else
        start = bef.next;
    ListNode SS = start;
    ListNode flag = null;
    ListNode next = start.next;
    for( int i = 0;i<n-m;i++){
        next = start.next;
        start.next = flag;
        flag = start;
        start = next;
    }
    ListNode end = start.next;
    start.next = flag;
    if( end == null && bef == null)
        return start;
    else if( end == null){
        bef.next = start;
        return head;
    }else if( bef == null){
        SS.next = end;
        return start;
    }else {
        bef.next = start;
        SS.next = end;
        return head;
    }

    }
}