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.

 ref: http://www.cnblogs.com/feiling/p/3263501.html

添加一个safeguard，防止处理m=1的情况，用p1, p2, p3记录m-1, m, n这几个节点的引用

当处理到m+1~n这些节点时将，将它们的next指向前一个节点

/**
 * 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 || m == n)
            return head;
        
        ListNode fake = new ListNode(Integer.MIN_VALUE);
        fake.next = head;
        head = fake;
        
        int count = 0;
        ListNode pre = null, p = head, p1 = null, p2 = null, p3 =null;
        while(count <= n){
            if(count == m-1){
                p1 = p;
            }else if(count == m){
                p2 = p;
            }else if(count == n){
                p3 = p;
            }
            
            ListNode tmp = p.next;
            
            if(count >= m+1 && count <= n){
                p.next = pre;
            }
            pre = p;
            p = tmp;
            count++;
        }
        p1.next = p3;
        p2.next = p;
        
        return fake.next;
    }
}