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.

相关问题是

Reverse Nodes in k-Group

public class Solution {
public ListNode reverseBetween(ListNode head, int m, int n) {
ListNode h = new ListNode(-1);
h.next = head;
ListNode pre = h;
int cnt = 1;
while(cnt<m){
pre = pre.next;
cnt++;
}
ListNode mnode = pre.next;
ListNode nnode = mnode;
while(cnt<n){
nnode = nnode.next;
cnt++;
}
ListNode end = nnode.next;
reverse(pre, end);
return h.next;
}

public void reverse(ListNode pre, ListNode end){ // B keep to being inserted after pre  pre-1-2-3-end, 1= A, 2=B, 3=t
ListNode A = pre.next, B  = A.next; // A 定义为B前面一位, 不断的把B塞到pre后面（也就是list的第一位，然后把原来B的前面A链接到B的后面t上去
while( B!=end){
ListNode t =  B.next;
B.next = pre.next;
pre.next = B;
A.next = t;
B=t;
}
}
}