题解 | #合并k个已排序的链表#

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode mergeKLists(ArrayList<ListNode> lists) {
        return mergeList(lists,0,lists.size()-1);
    }
    public ListNode mergeList(List<ListNode> lists,int left,int right){
        if(left == right){
            return lists.get(left);
        }
        if(left > right){
            return null;
        }
        int mid = left + (right - left)/2;
        return merge(mergeList(lists,left,mid),mergeList(lists,mid+1,right));
    }
    public ListNode merge(ListNode l1,ListNode l2){
        if(l1 == null){
            return l2;
        }
        if(l2 == null){
            return l1;
        }
        ListNode head = new ListNode(-1);
        ListNode node = head;
        while(l1 != null && l2 != null){
            if(l1.val < l2.val){
                node.next = l1;
                l1 = l1.next;
            }else{
                node.next = l2;
                l2 = l2.next;
            }
            node = node.next;
        }
        if(l1 != null){
            node.next = l1;
        }else{
           node.next = l2;
        }
        return head.next;
    }
}