☆打卡算法☆LeetCode 23、合并K个升序链表 算法解析

示例 1：
输入：lists = [[1,4,5],[1,3,4],[2,6]]
输出：[1,1,2,3,4,4,5,6]
解释：链表数组如下：
[
  1->4->5,
  1->3->4,
  2->6
]
将它们合并到一个有序链表中得到。
1->1->2->3->4->4->5->6

示例 2：
输入： lists = [[]]
输出： []

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int val=0, ListNode next=null) {
 *         this.val = val;
 *         this.next = next;
 *     }
 * }
 */
public class Solution {
    public ListNode MergeKLists(ListNode[] lists) {
        if(lists == null || lists.Length == 0) {
            return null;
        }

        return MergeKLists(lists, 0, lists.Length-1);
    }

    private ListNode MergeKLists(ListNode[] lists, int left, int right) {
        if(left == right) {
            return lists[left];
        }
        int mid = left + (right-left)/2;
        ListNode l1 = MergeKLists(lists, left, mid);
        ListNode l2 = MergeKLists(lists, mid+1, right);

        return MergeTwoLists(l1, l2);
    }

    private ListNode MergeTwoLists(ListNode l1, ListNode l2) {
        if(l1 == null) {
            return l2;
        }
        if(l2 == null) {
            return l1;
        }

        ListNode dummy = new ListNode();
        ListNode p1 = l1, p2 = l2, cur = dummy;
        while(p1 != null && p2 != null) {
            if(p1.val < p2.val) {
                cur.next = p1;
                cur = cur.next;
                p1 = p1.next;
            } else {
                cur.next = p2;
                cur = cur.next;
                p2 = p2.next;
            }
        }

        cur.next = p1 == null ? p2 : p1;

        return dummy.next;
    }
}