Description

Reverse a linked list from position m to n. Do it in one-pass.

Note: 1 ≤ m ≤ n ≤ length of list.

Example:

1 2	Input: 1->2->3->4->5->NULL, m = 2, n = 4 Output: 1->4->3->2->5->NULL

解法

基本的链表操作题，为解决对于链表头的处理，常在表头插入空项，即dummy指针。对于翻转链表，采用在链表头部插入的技术即可。

具体代码如下：

class Solution {
    public ListNode reverseBetween(ListNode head, int m, int n) {
        if (m == n || head == null) {
            return head;
        }
        ListNode dummy = new ListNode(-1);
        dummy.next = head;
        int cnt = 1;
        ListNode pre = dummy;
        ListNode next;
        ListNode tmp = head;
        ListNode inserthead;
        for (; cnt < m; cnt++) {
            pre = tmp;
            tmp = pre.next;
        }
        inserthead = tmp;
        next = tmp.next;
        inserthead.next = null;
        ListNode tail = inserthead;
        for (; cnt < n; cnt++) {
            tmp = next;
            next = next.next;
            tmp.next = inserthead;
            inserthead = tmp;
        }
        tail.next = next;
        pre.next = inserthead;
        return dummy.next;
    }
}