23. 合并K个排序链表

输入:
[
  1->4->5,
  1->3->4,
  2->6
]
输出: 1->1->2->3->4->4->5->6

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    private ListNode merge(ListNode l1, ListNode l2) {
        ListNode l = new ListNode(0);
        ListNode p = l;

        while (l1 != null && l2 != null) {
            if (l1.val < l2.val) {
                p.next = l1;
                l1 = l1.next;
            } else {
                p.next = l2;
                l2 = l2.next;
            }
            p = p.next;
        }

        if (l1 != null)
            p.next = l1;

        if (l2 != null)
            p.next = l2;

        return l.next;
    }

    public ListNode mergeKLists(ListNode[] lists) {
        return partion(lists, 0, lists.length - 1);
    }

    private ListNode partion(ListNode[] lists, int s, int e) {
        if (s == e) {
            return lists[s];
        }
        if (s < e) {
            int q = (s + e) / 2;
            ListNode l1 = partion(lists, s, q);
            ListNode l2 = partion(lists, q + 1, e);
            return merge(l1, l2);
        } else
            return null;
    }
    
    private ListNode merge(ListNode l1, ListNode l2) {
        ListNode l = new ListNode(0);
        ListNode p = l;

        while (l1 != null && l2 != null) {
            if (l1.val < l2.val) {
                p.next = l1;
                l1 = l1.next;
            } else {
                p.next = l2;
                l2 = l2.next;
            }
            p = p.next;
        }

        if (l1 != null)
            p.next = l1;

        if (l2 != null)
            p.next = l2;

        return l.next;
    }
}