题解 | #合并k个已排序的链表#

import java.util.*;

/*
 * public class ListNode {
 *   int val;
 *   ListNode next = null;
 *   public ListNode(int val) {
 *     this.val = val;
 *   }
 * }
 */

public class Solution {
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param lists ListNode类ArrayList 
     * @return ListNode类
     */
    public ListNode mergeTwo(ListNode pHead1, ListNode pHead2){
        ListNode head = new ListNode(-1);
        ListNode cur = head;
        while(pHead1 != null && pHead2 != null){
            if(pHead1.val <= pHead2.val){
                cur.next = pHead1;
                pHead1 = pHead1.next;
            }else{
                cur.next = pHead2;
                pHead2 =  pHead2.next;
            }
            cur = cur.next;
        }
        if(pHead1 != null){
            cur.next = pHead1;
        }else{
            cur.next = pHead2;
        }
        return head.next;
    }

    public ListNode merge(ArrayList<ListNode> lists, int left, int right){
        if(left > right){
            return null;
        }
        if(left == right){ 
            return lists.get(left);
        }
        int mid = (left + right)/2;
        //对半划分，将划分后的子问题合并成新的链表
        return mergeTwo(merge(lists,left,mid), merge(lists,mid + 1,right));
    }

    public ListNode mergeKLists (ArrayList<ListNode> lists) {
        // write code here
        return merge(lists,0,lists.size()-1);
    }
}