23. Merge k Sorted Lists (JAVA)

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

Example:

Input:
[
  1->4->5,
  1->3->4,
  2->6
]
Output: 1->1->2->3->4->4->5->6

 

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        if(lists == null || lists.length == 0)  //pay attention to null
            return null;

        ListNode head = null;
        ListNode cur = head; //cur.next point to the node to give value
        PriorityQueue<ListNode> minHeap = new PriorityQueue<>(lists.length,
            new Comparator<ListNode>(){ //小顶堆，默认容量11，这里改为列表长度
            @Override
            public int compare(ListNode n1, ListNode n2){
                return n1.val - n2.val;
            }
        });

        for(ListNode n: lists){
            if(n!=null) minHeap.add(n); //pay attention to null
        }

        while(minHeap.size()>0){
            //pop
            if(head==null){
                head = minHeap.poll();
                cur = head;
            }
            else{
                cur.next = minHeap.poll();
                cur = cur.next;
            }

            //push
            if(cur.next != null){
                minHeap.add(cur.next);
            }
        }
        return head;
    }
}

PriorityQueue是通过小顶堆实现的，如果要使用大顶堆，那么需要自定义Comparator函数。

List问题特别注意：需要额外讨论null的情况，比如本题中，需要保证PriorityQueue不能加入null元素。