题解 | 合并k个已排序的链表

基于合并两个已排序的链表，将k个已排序的链表不断的划分为2组，直到划分成2个或1个链表为止

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 mergeKLists (ArrayList<ListNode> lists) {
        // write code here
        int n = lists.size();
        if(n == 0) return null;

        if(n == 2) return Merge(lists.get(0), lists.get(1));
        else if(n == 1) return lists.get(0);
        else {
            int half = n / 2;   // 分治
            ListNode left = mergeKLists(new ArrayList(lists.subList(0, half)));	// subList的参数遵循左闭右开的原则
            ListNode right = mergeKLists(new ArrayList(lists.subList(half, n)));

            return Merge(left, right);
        }
    }

    /**
     * 合并两个已排序的链表
     */
     public static ListNode Merge(ListNode phead1, ListNode phead2) {
        ListNode dummy = new ListNode(-1);
        ListNode head = dummy;
        while(phead1 != null && phead2 != null) {
            if(phead1.val <= phead2.val) {
                head.next = phead1;
                phead1 = phead1.next;
            } else {
                head.next = phead2;
                phead2 = phead2.next;
            }
            head = head.next;
        }

        if(phead1 == null) head.next = phead2;
        if(phead2 == null) head.next = phead1;
        return dummy.next;
     }
}