鏈表merge-k-sorted-lists-leetcode練習題

import java.util.Comparator;
import java.util.PriorityQueue;

/**
 * 合併 k 個排序鏈表，返回合併後的排序鏈表。請分析和描述算法的複雜度。
 * <p>
 * 示例:
 * <p>
 * 輸入:
 * [
 * 1->4->5,
 * 1->3->4,
 * 2->6
 * ]
 * 輸出: 1->1->2->3->4->4->5->6
 * <p>
 * 來源：力扣（LeetCode）
 * 鏈接：https://leetcode-cn.com/problems/merge-k-sorted-lists
 * 著作權歸領釦網絡所有。商業轉載請聯繫官方授權，非商業轉載請註明出處。
 */
public class Solution {
       /**
     * 優先級隊列求解
     *
     * @param lists
     * @return
     */
    public ListNode mergeKLists(ListNode[] lists) {
        if (lists == null || lists.length == 0) return null;
        PriorityQueue<ListNode> queue = new PriorityQueue<>(lists.length, (o1, o2) -> {
            if (o1.val < o2.val) return -1;
            else if (o1.val == o2.val) return 0;
            else return 1;
        });
        ListNode dummy = new ListNode(0);
        ListNode p = dummy;
        for (ListNode node : lists) {
            if (node != null) queue.add(node);
        }
        while (!queue.isEmpty()) {
            p.next = queue.poll();
            p = p.next;
            if (p.next != null) queue.add(p.next);
        }
        return dummy.next;
    }

    /**
     * 歸併排序思路
     *
     * @param lists
     * @return
     */
    public ListNode mergeKLists2(ListNode[] lists) {
        if (lists == null || lists.length == 0) return null;
        return merge(lists, 0, lists.length - 1);
    }

    private ListNode merge(ListNode[] lists, int left, int right) {
        if (left == right) return lists[left];
        int mid = left + (right - left) / 2;
        ListNode l1 = merge(lists, left, mid);
        ListNode l2 = merge(lists, mid + 1, right);
        return mergeTwoLists(l1, l2);
    }

    private ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if (l1 == null) return l2;
        if (l2 == null) return l1;
        if (l1.val < l2.val) {
            l1.next = mergeTwoLists(l1.next, l2);
            return l1;
        } else {
            l2.next = mergeTwoLists(l1, l2.next);
            return l2;
        }
    }

}