Reverse Linked List II

描述
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->nullptr, m = 2 and n = 4,
return 1->4->3->2->5->nullptr.
Note: Given m, n satisfy the following condition: 1 ≤ m ≤ n ≤ length of list.
分析

这题非常繁琐，有很多边界检查，15 分钟内做到 bug free 很有难度！

代码

public class ReverseLinkedList {

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        ListNode head = new ListNode(1);
        // head.add(1);
        head.add(2);
        head.add(3);
        head.add(4);
        head.add(5);
        head.print();
        System.out.println();
        reverseBetween(head, 2, 4).print();
    }

    public static ListNode reverseBetween(ListNode head, int m, int n) {
        ListNode newhead = new ListNode(-1);
        newhead.next = head;
        if (head == null || head.next == null)
            return newhead.next;

        ListNode startpoint = newhead;
        ListNode node1 = null;
        ListNode node2 = null;
        for (int i = 0; i < n; i++) {
            if (i < m - 1) {
                startpoint = startpoint.next;// 寻找真正的startpoint m-1的位置
            } else if (i == m - 1) { // 开始变换的第一轮，找到初始位置
                node1 = startpoint.next;
                node2 = node1.next;
            } else {
                node1.next = node2.next; // node1交换到node2后面
                node2.next = startpoint.next; // node2连startpoint后面的数，每次都移到最前
                startpoint.next = node2; // startpoint连node2
                node2 = node1.next; // node2移到node1后面，进行下一次的交换

            }
        }
        return newhead.next;
    }
}