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.

Reverse Linked List II

/**
 * 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) {
        
        if (head == null || head.next == null)
            return head;
        
        if (m >= n || m < 0 || n < 0)
            return head;
        
        ListNode fakeHead = new ListNode(0);
        fakeHead.next = head;
        ListNode p1 = fakeHead;
        ListNode hd = p1;
        ListNode tl = p1;
        int count = 0;
        
        while (p1 != null) {
            
            if (count+1 == m) {
                hd = p1;
            }
            
            if (count == n) {
                tl = p1.next;
                break;
            }
            count++;
            p1 = p1.next;
        }
        
        hd = reverseList(hd, tl);
        
        return fakeHead.next;
    }
    
    //reverse the list between head and tail, not include head and tail
    public ListNode reverseList(ListNode head, ListNode tail) {
        
        ListNode curr = head.next;
        ListNode prev = tail;
        
        while (curr != tail) {
            ListNode temp = curr.next;
            curr.next = prev;
            prev = curr;
            curr = temp;
        }
        head.next = prev;
    
        return head;
    }
}