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;
}
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);
}
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);
}
public Container(ListNode list)
{
this.List = list;
}
