标签:通过 lists 注意 attention def put scribe tno 定义
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。
23. Merge k Sorted Lists (JAVA)
标签:通过 lists 注意 attention def put scribe tno 定义
原文地址:https://www.cnblogs.com/qionglouyuyu/p/10775700.html