LeetCode

题目：

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：

以下代码很多，但逻辑很直接，应该还可以简化

package list;

public class ReverseLinkedListII {

    public ListNode reverseBetween(ListNode head, int m, int n) {
        if (m == n) return head;
        ListNode p = new ListNode(0);
        p.next = head;
        ListNode firstPivot = null;
        ListNode firstPivotNext = null;
        ListNode prev = p;
        ListNode node = head;
        int i = 1;
        while (i <= n) {
            if (i > m && i < n) {
                ListNode tmp = node;
                node = node.next;
                tmp.next = prev;
                prev = tmp;
            } else if (i < m) {
                node = node.next;
                prev = prev.next;
            } else if (i == m) {
                firstPivot = prev;
                firstPivotNext = node;
                ListNode tmp = node;
                node = node.next;
                tmp.next = null;
                prev = tmp;
            } else if (i == n) {
                ListNode tmp = node;
                node = node.next;
                tmp.next = prev;
                firstPivot.next = tmp;
                firstPivotNext.next = node;
            }
            ++i;            
        }
        return p.next;
    }
    
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        ListNode a1 = new ListNode(1);
        ListNode a2 = new ListNode(2);
        ListNode a3 = new ListNode(3);
        ListNode a4 = new ListNode(4);
        ListNode a5 = new ListNode(5);
        a1.next = a2;
        a2.next = a3;
        a3.next = a4;
        a4.next = a5;
        a5.next = null;
        ReverseLinkedListII r = new ReverseLinkedListII();
        ListNode head = r.reverseBetween(a1, 2, 5);
        while (head != null) {
            System.out.println(head.val);
            head = head.next;
        }
    }

}