leetcode刷題-23.Merge K Sorted Lists

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

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

合併k個有序列表,我採用的是優先級隊列(PriorityQueue),將所有列表放入PriorityQueue中，然後類似寬度優先搜索，取出最小的，如果該鏈表的next不爲空，將next加入到優先級隊列中

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    private Comparator<ListNode> ListNodeComporator = new Comparator<ListNode>() {
        public int compare(ListNode left, ListNode right) {
            return left.val - right.val;
        }
    };
    
    public ListNode mergeKLists(ListNode[] lists) {
    	//邊界處理
        if (lists == null || lists.length == 0) {
            return null;
        }
        //將List放入PriorityQueue
        Queue<ListNode> heap = new PriorityQueue<>(lists.length, ListNodeComporator);
        for (int i = 0; i < lists.length; i++) {
            if (lists[i] != null) {
                heap.offer(lists[i]);
            }
        }
        
        ListNode dummy = new ListNode(0);
        ListNode tail = dummy;
        
        while(!heap.isEmpty()) {
        	//取出當前最小
            ListNode node = heap.poll();
            //構建結果列表
            tail.next = node;
            tail = tail.next;
            //將最小node的next加入PriorityQueue
            if (node.next != null) {
                heap.offer(node.next);
            }
        }
        return dummy.next;
    }
}