LeetCode_23.合并K个升序链表

给你一个链表数组，每个链表都已经按升序排列。

请你将所有链表合并到一个升序链表中，返回合并后的链表。

 

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：[]

示例 3：

输入：lists = [[]]
输出：[]

 

提示：

• k == lists.length
• 0 <= k <= 10^4
• 0 <= lists[i].length <= 500
• -10^4 <= lists[i][j] <= 10^4
• lists[i] 按 升序 排列
• lists[i].length 的总和不超过 10^4

C#代码

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode MergeKLists(ListNode[] lists)
    {
        int index = 0;
        int count = lists.Length;
        int temp = 1;

        if (count == 0) return null;

        while (temp < count)
        {
            index = 0;
            while (index + temp < count)
            {
                lists[index] = MergeTwoList(lists[index], lists[index + temp]);
                index += temp * 2;
            }
            temp *= 2;
        }
        return lists[0];
    }

    public ListNode MergeTwoList(ListNode node1, ListNode node2)
    {
        Stack<int> stack = new Stack<int>();
        int val1, val2;
        while (node1 != null && node2 != null)
        {
            val1 = node1.val;
            val2 = node2.val;
            if (val1 <= val2)
            {
                stack.Push(val1);
                node1 = node1.next;
            }
            else
            {
                stack.Push(val2);
                node2 = node2.next;
            }
        }

        while (node1 != null)
        {
            val1 = node1.val;
            stack.Push(val1);
            node1 = node1.next;
        }

        while (node2 != null)
        {
            val2 = node2.val;
            stack.Push(val2);
            node2 = node2.next;
        }

        ListNode node = null;
        while (stack.Count > 0)
        {
            node = new ListNode() { val = stack.Pop(), next = node };
        }
        return node;
    }
}