public static ListNode merge2list(ListNode head1, ListNode head2)
        {
            ListNode node = new ListNode(0),
                     res  = node;

            while(head1 != null && head2 != null )
            {
                if(head1.val <= head2.val)
                {
                    res.next = head1;
                    head1 = head1.next;
                }
                else
                {
                    res.next = head2;
                    head2 = head2.next;
                }

                res = res.next;   // it is delayed to reset to next one, after head1, head2 resets to its next.
            }

               if(head1 != null)
                res.next = head1;
               else if(head2 != null)
                res.next = head2;

               return node.next;
        }
        /*
         * Leetcode 23: Merge K Sorted Lists
        Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.

        合并k个有序的链表,我们假设每个链表的平均长度是n。这一题需要用到合并两个有序的链表子过程
         */
        /*
         * http://www.cnblogs.com/TenosDoIt/p/3673188.html
         * 算法1: Naive solution

            最傻的做法就是先1、2合并,12结果和3合并,123结果和4合并,…,123..k-1结果和k合并,我们计算一下复杂度。

            1、2合并,遍历2n个节点

            12结果和3合并,遍历3n个节点

            123结果和4合并,遍历4n个节点

            …

            123..k-1结果和k合并,遍历kn个节点

            总共遍历的节点数目为n(2+3+…+k) = n*(k^2+k-2)/2, 因此时间复杂度是O(n*(k^2+k-2)/2) = O(nk^2),代码如下:
         */
        public static ListNode mergeKLists(ListNode[] lists)
        {
            if(lists.Length == 0)
                return null;

            ListNode res = lists[0];

            for(int i = 1; i < lists.Length; i++)
                res = merge2list(res, lists[i]);

            return res;
        }
예제 #3
0
            public ListNode MergeKLists(ListNode[] lists)
            {
                var q = new PriorityQueue <Container>();

                foreach (var list in lists)
                {
                    if (list == null)
                    {
                        continue;
                    }
                    q.Enqueue(new Container(list));
                }
                ListNode  retval = null, current = null;
                Container min;
                ListNode  listToInsert;

                while (q.Count() > 0)
                {
                    min          = q.Dequeue();
                    listToInsert = min.List.next;
                    if (retval == null)
                    {
                        retval  = new ListNode(min.Element);
                        current = retval;
                    }
                    else
                    {
                        current.next = new ListNode(min.Element);
                        current      = current.next;
                    }
                    if (listToInsert != null)
                    {
                        q.Enqueue(new Container(listToInsert));
                    }
                }
                return(retval);
            }
예제 #4
0
 public ListNode(int val = 0, ListNode next = null)
 {
     this.val  = val;
     this.next = next;
 }
        static void Main(string[] args)
        {
            ListNode l1 = new ListNode(1);
            l1.next = new ListNode(4);
            l1.next.next = new ListNode(7);

            ListNode l2 = new ListNode(2);
            l2.next = new ListNode(5);
            l2.next.next = new ListNode(8);

            ListNode l3 = new ListNode(3);
            l3.next = new ListNode(6);

            ListNode[] lists = new ListNode[3]{l1, l2, l3};
            ListNode test = mergeKLists(lists);
        }
예제 #6
0
 public Container(ListNode list)
 {
     this.List = list;
 }