23. merge k sorted list

Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.

解法1: PriorityQueue, O(NlogM)

M是queue的长度,N是所有node的总数。
一个简单的PriorityQueue的应用。

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* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class {
public ListNode mergeKLists(ListNode[] lists) {
if (lists == null || lists.length == 0) {
return null;
}
ListNode dummy = new ListNode(0);
ListNode tail = dummy;
Queue<ListNode> queue = new PriorityQueue<>((a,b) -> a.val - b.val);
for (int i = 0; i < lists.length; i++) {
if (lists[i] != null) {
queue.offer(lists[i]);
}
}
while (!queue.isEmpty()) {
ListNode next = queue.poll();
tail.next = next;
tail = tail.next;
// add another node if exist
if (next.next != null) {
queue.offer(next.next);
}
}
return dummy.next;
}
}