leetcode--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./**
* 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(n==m) return head;
if(m>n){
int t = n;
n = m;
m = t;
}
ListNode h = head;
ListNode res = new ListNode(-1);
ListNode pre = res;
pre.next = h;
int count = 1;
//pre为m前的值
while(h!=null&&count<m){
pre = pre.next;
h = h.next;
count++;
}
ListNode cur = h;//找到m对应的值
ListNode fhead = new ListNode(-1);
ListNode last = cur;//第一个，然后会成为最后一个
while(count<=n){
pre.next = cur.next;
cur.next = fhead.next;
fhead.next = cur;
cur = pre.next;
count++;
}
last.next = pre.next;
pre.next = fhead.next;
return res.next;
}
}