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;
* next = null;
* }
* }
*/
public class Solution {
public ListNode reverseBetween(ListNode head, int m, int n) {
ListNode mPrev=null,mNode=head;
ListNode nNode=head,nNext=null;
for(int i=1;i<m;i++){
mPrev=mNode;
mNode=mNode.next;
}
for(int i=1;i<n;i++){
nNode=nNode.next;
}
nNext=nNode.next;
nNode.next=null;
if(mPrev!=null) mPrev.next=nNode;//
ListNode tail=mNode;
ListNode rHead=reverse(mNode);
if(tail!=null) tail.next=nNext;
if(head==tail) return rHead;
else return head;
}
public ListNode reverse(ListNode head){
ListNode prev=null;
ListNode node=head;
ListNode newHead=null;
while(node!=null){
ListNode next=node.next;
if(next==null){
newHead=node;
}
node.next=prev;
prev=node;
node=next;
}
return newHead;
}
}