1.题目描述
Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.
合并 k 个排序链表，返回合并后的排序链表。请分析和描述算法的复杂度。
Example:
Input:
[
1->4->5,
1->3->4,
2->6
]
Output: 1->1->2->3->4->4->5->6
2.Solutions
迭代、分治版本：
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public static ListNode (ListNode[] lists){
return partion(lists,0,lists.length-1);
}
public static ListNode partion(ListNode[] lists,int s,int e){
if(s==e) return lists[s];
if(s<e){
int q=s+(e-s)/2;
ListNode l1=partion(lists,s,q);
ListNode l2=partion(lists,q+1,e);
return merge(l1,l2);
}else
return null;
}
public static ListNode merge(ListNode l1,ListNode l2){
if(l1==null) return l2;
if(l2==null) return l1;
if(l1.val<l2.val){
l1.next=merge(l1.next,l2);
return l1;
}else{
l2.next=merge(l1,l2.next);
return l2;
}
}
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优先队列：
If someone understand how priority queue works, then it would be trivial to walk through the codes.
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public ListNode (List<ListNode> lists) {
if (lists==null||lists.size()==0) return null;
PriorityQueue<ListNode> queue= new PriorityQueue<ListNode>(lists.size(),new Comparator<ListNode>(){
public int compare(ListNode o1,ListNode o2){
if (o1.val<o2.val)
return -1;
else if (o1.val==o2.val)
return 0;
else
return 1;
}
});
ListNode dummy = new ListNode(0);
ListNode tail=dummy;
for (ListNode node:lists)
if (node!=null)
queue.add(node);
while (!queue.isEmpty()){
tail.next=queue.poll();
tail=tail.next;
if (tail.next!=null)
queue.add(tail.next);
}
return dummy.next;
}
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（完）
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