题解 | #合并k个已排序的链表#

import java.util.ArrayList;
public class Solution {
    //两个链表合并 fast-template
    public ListNode Merge(ListNode list1, ListNode list2) {
        //一个已经为空了，直接返回另一个
        if (list1 == null)
            return list2;
        if (list2 == null)
            return list1;
        //加一个表头
        ListNode head = new ListNode(0);
        ListNode cur = head;
        //两个链表都要不为空
        while (list1 != null && list2 != null) {
            //取较小值的结点
            if (list1.val <= list2.val) {
                cur.next = list1;
                //只移动取值的指针
                list1 = list1.next;
            } else {
                cur.next = list2;
                //只移动取值的指针
                list2 = list2.next;
            }
            //指针后移
            cur = cur.next;
        }
        //哪个链表还有剩，直接连在后面
        if (list1 != null)
            cur.next = list1;
        else
            cur.next = list2;
        //返回值去掉表头
        return head.next;
    }
    //划分合并区间
    ListNode divideMerge(ArrayList<ListNode> lists, int left, int right) {
        if (left > right)
            return null;
        //中间一个的情况
        else if (left == right)
            return lists.get(left);
        //从中间分成两段，再将合并好的两段合并
        int mid = (left + right) / 2;
        return Merge(divideMerge(lists, left, mid), divideMerge(lists, mid + 1, right));
    }
    public ListNode mergeKLists(ArrayList<ListNode> lists) {
        return divideMerge(lists, 0, lists.size() - 1);
    }
}