You are given an array of k linked-lists lists, each linked-list is sorted in ascending order.
Merge all the linked-lists into one sorted linked-list and return it.
Example 1:
Input: lists = [[1,4,5],[1,3,4],[2,6]]
Output: [1,1,2,3,4,4,5,6]
Explanation: The linked-lists are:
[
1->4->5,
1->3->4,
2->6
]
merging them into one sorted list:
1->1->2->3->4->4->5->6
Example 2:
Input: lists = []
Output: []
Example 3:
Input: lists = [[]] Output: []
就是用q啊
/**
* 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.
*/
//comparator
private Comparator<ListNode> ListNodeComparator = new Comparator<ListNode>() {
public int compare(ListNode left, ListNode right) {
return left.val - right.val;
}
};
public ListNode mergeKLists(ListNode[] lists) {
//new heap
if (lists.length == 0 || lists == null) {
return null;
}
Queue<ListNode> heap = new PriorityQueue<ListNode>(lists.length,ListNodeComparator);
for (int i = 0; i < lists.length; i++) {
if (lists[i] != null) {
//先加每个的head，1 1 2,然后小的才有资格继续加 4 3
//加进来之后一起去和2比较，2更小，然后继续加
//把整个的list加进去
heap.add(lists[i]);
}
}
//add to the lists
ListNode dummy = new ListNode(0);
ListNode tail = dummy;
while(!heap.isEmpty()) {
ListNode head = heap.poll();
System.out.println("head = " + head.val);
tail.next = head;
tail = head;
if (head.next != null) {
System.out.println("添加的head.next = " + head.next.val);
System.out.println(" ");
heap.add(head.next);
}
}
return dummy.next;
}
}