23. 合并K个升序链表

23. 合并K个升序链表

思路：

　　将两个有序链表合并，推广到n个链表的合并，就是将n个链表两两合并，再合并，就是归并排序的思维，现将数组递归到单个元素终止，然后执行两两merge。

https://leetcode-cn.com/problems/merge-k-sorted-lists/solution/he-bing-kge-pai-xu-lian-biao-by-leetcode-solutio-2/

/**
 * 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) {
        if (lists == null || lists.length == 0) {
            return null;
        }
        return merge(lists, 0, lists.length - 1);
    }

    public ListNode merge(ListNode[] lists, int l, int r) {
        if (l == r) {
            return lists[l];
        }
        if (l > r) {
            return null;
        }
        int mid = (l + r) >> 1;
        return mergeTwoListNode(merge(lists, l, mid), merge(lists, mid + 1, r));
    }

    // 合并两个链表
    public ListNode mergeTwoListNode(ListNode listNode1, ListNode listNode2) {
        ListNode head = new ListNode(-1);
        ListNode result=head;
        while (listNode1 != null && listNode2 != null) {
            if(listNode1.val<listNode2.val){
                result.next=listNode1;
                listNode1=listNode1.next;
            }else {
                result.next=listNode2;
                listNode2=listNode2.next;
            }
            result=result.next;
        }
        if(listNode1!=null){
            result.next=listNode1;
        }else {
            result.next=listNode2;
        }
        return head.next;
    }
}

。。