Merge?k?sorted linked lists and return it as one sorted list. Analyze and describe its complexity.
?
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode mergeKLists(ListNode[] lists) {
if (lists==null || lists.length==0) {
return null;
}
PriorityQueue<ListNode> queue = new PriorityQueue<ListNode>(lists.length, new Comparator<ListNode>() {
@Override
public int compare(ListNode o1, ListNode o2) {
if (o1.val < o2.val) {
return -1;
} else if (o1.val > o2.val) {
return 1;
} else {
return 0;
}
}
});
for (ListNode list : lists) {
if (list == null) {
continue;
}
queue.add(list);
}
ListNode head = new ListNode(0);
ListNode curNode = head;
while (!queue.isEmpty()) {
ListNode t = queue.poll();
curNode.next = t;
curNode = t;
if (t.next != null) {
queue.add(t.next);
}
}
return head.next;
}
}
?
原文:http://hcx2013.iteye.com/blog/2216801