Merge k Sorted Lists

Merge k sorted linked lists and return it as one sorted list.

Analyze and describe its complexity.

Example

Given lists:

[
2->4->null,
null,
-1->null
],

return

-1->2->4->null

.

分析：

利用heap来保存数组中每一个list的第一个node, 当取出一个node以后，再把那个node的next node加进来，直到所有node都已被加进来。

/**
* Definition for ListNode.
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int val) {
*         this.val = val;
*         this.next = null;
*     }
* }
*/
public class Solution {
/**
* @param lists: a list of ListNode
* @return: The head of one sorted list.
*/
public ListNode mergeKLists(List<ListNode> lists) {
PriorityQueue<ListNode> heap = new PriorityQueue<ListNode>(10,new Comparator<ListNode>(){
@Override
public int compare(ListNode n1, ListNode n2) {
return n1.val - n2.val;
}
});

for(int i = 0; i < lists.size(); i++) {
ListNode node = lists.get(i);
if(node != null) {
heap.add(node);
}
}

ListNode head = null;
ListNode cur = null;
while(heap.size() > 0) {
ListNode node = heap.poll();
if(head == null) {
head = node;
}

if (cur != null) {
cur.next = node;
}

cur = node;

if(node.next != null) {
heap.add(node.next);
}
}
return head;
}
}