23. 合并 K 个升序链表（递归分治）

这是我的第一个自己ak的分治题目！！！好耶！！（骄傲脸

思路参考：148. 排序链表（归并排序）

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        return dfs(lists, 0, lists.length - 1);
    }
    public ListNode dfs(ListNode[] lists, int head, int tail) {
        if (head > tail) return null;
        if (head == tail) return lists[head];
        int mid = (head + tail) / 2;
        ListNode node1 = dfs(lists, head, mid);
        ListNode node2 = dfs(lists, mid + 1, tail);
        ListNode node = merge(node1, node2);
        return node;
    }
    public ListNode merge(ListNode node1, ListNode node2) {
        ListNode node = new ListNode(0);
        ListNode temp = node, temp1 = node1, temp2 = node2;
        while (temp1 != null && temp2 != null) {
            if (temp1.val < temp2.val) {
                temp.next = temp1;
                temp1 = temp1.next;
            } else {
                temp.next = temp2;
                temp2 = temp2.next;
            }
            temp = temp.next;
        }
        if (temp1 != null) temp.next = temp1;
        if (temp2 != null) temp.next = temp2;
        return node.next;
    }
}