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.

思路： We set 2 pointers to count the m and n . If we find the m, then we get a pre pointer before m. Then we cut the list off and insert the rest nodes to the pre.next by using a tail insertion. When we find the mth node and get the m.next, we need to find the tail of the front list and the tail.next node is m.next.

易错点： 1. 注意找到m 之后， pre.next = null 来断开list. 
2. 注意尾插的时候， nextNode 要先走到下一位 再 插入。
3. 最后末尾别忘记把两个list 合成一个。

/**
 * 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 newHead = new ListNode(0);
        ListNode pre = newHead;
        ListNode nextNode = head;
        newHead.next = head;
        int count = 1;
        while(count < m ){// 找到第一个 m
            count++;
            pre = pre.next;
            nextNode = nextNode.next;
        }
        pre.next = null;
        while(count <= n ){// 每次尾插一个节点， 即为逆序
            ListNode cur = nextNode;
            nextNode = nextNode.next;
            cur.next = pre.next;
            pre.next = cur;
            count++;
        }
        ListNode cur = pre;
        while(cur.next != null){// 找到前半部分List  的 末尾
            cur = cur.next;
        }
        cur.next = nextNode;// 连接起来
        return newHead.next;
    }
}