leetcode-优先级队列-23

/**
 * <p>给你一个链表数组，每个链表都已经按升序排列。</p>
 *
 * <p>请你将所有链表合并到一个升序链表中，返回合并后的链表。</p>
 *
 * <p>&nbsp;</p>
 *
 * <p><strong>示例 1：</strong></p>
 *
 * <pre><strong>输入：</strong>lists = [[1,4,5],[1,3,4],[2,6]]
 * <strong>输出：</strong>[1,1,2,3,4,4,5,6]
 * <strong>解释：</strong>链表数组如下：
 * [
 * 1-&gt;4-&gt;5,
 * 1-&gt;3-&gt;4,
 * 2-&gt;6
 * ]
 * 将它们合并到一个有序链表中得到。
 * 1-&gt;1-&gt;2-&gt;3-&gt;4-&gt;4-&gt;5-&gt;6
 * </pre>
 *
 * <p><strong>示例 2：</strong></p>
 *
 * <pre><strong>输入：</strong>lists = []
 * <strong>输出：</strong>[]
 * </pre>
 *
 * <p><strong>示例 3：</strong></p>
 *
 * <pre><strong>输入：</strong>lists = [[]]
 * <strong>输出：</strong>[]
 * </pre>
 *
 * <p>&nbsp;</p>
 *
 * <p><strong>提示：</strong></p>
 *
 * <ul>
 * <li><code>k == lists.length</code></li>
 * <li><code>0 &lt;= k &lt;= 10^4</code></li>
 * <li><code>0 &lt;= lists[i].length &lt;= 500</code></li>
 * <li><code>-10^4 &lt;= lists[i][j] &lt;= 10^4</code></li>
 * <li><code>lists[i]</code> 按 <strong>升序</strong> 排列</li>
 * <li><code>lists[i].length</code> 的总和不超过 <code>10^4</code></li>
 * </ul>
 * <div><div>Related Topics</div><div><li>链表</li><li>分治</li><li>堆（优先队列）</li><li>归并排序</li></div></div><br><div><li>👍 1874</li><li>👎 0</li></div>
 */

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import javax.xml.transform.Templates;
import java.util.Comparator;
import java.util.PriorityQueue;

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

        PriorityQueue<ListNode> pq = new PriorityQueue<>(new Comparator<ListNode>() {
            @Override
            public int compare(ListNode o1, ListNode o2) {
                return o1.val - o2.val;
            }
        });


        ListNode dummy = new ListNode(-1);
        ListNode curr = dummy;
        for (ListNode head : lists) {
            if (head != null) {
                pq.add(head);
            }
        }
        System.out.println("头节点都已放入");

        while (!pq.isEmpty()) {
            ListNode node = pq.poll();
            curr.next = node;
            curr =curr.next;

            if(node.next!=null){
                pq.add(node.next);
            }

        }

        return dummy.next;
    }
}
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