Leetcode-23-合并K个升序链表(Merge k Sorted Lists)

Leetcode-23-合并K个升序链表(Merge k Sorted Lists)

23. 合并K个升序链表

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给你一个链表数组，每个链表都已经按升序排列。

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。

示例 1：

输入：lists = [[1,4,5],[1,3,4],[2,6]]
输出：[1,1,2,3,4,4,5,6]
解释：链表数组如下：
[
  1->4->5,
  1->3->4,
  2->6
]
将它们合并到一个有序链表中得到。
1->1->2->3->4->4->5->6

示例 2：

输入：lists = []
输出：[]

示例 3：

输入：lists = [[]]
输出：[]

提示：

• k == lists.length
• 0 <= k <= 10^4
• 0 <= lists[i].length <= 500
• -10^4 <= lists[i][j] <= 10^4
• lists[i] 按 升序 排列
• lists[i].length 的总和不超过 10^4

---

23. Merge k Sorted Lists

Hard

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You are given an array of k linked-lists lists, each linked-list is sorted in ascending order.

Merge all the linked-lists into one sorted linked-list and return it.

Example 1:

Input: lists = [[1,4,5],[1,3,4],[2,6]]
Output: [1,1,2,3,4,4,5,6]
Explanation: The linked-lists are:
[
  1->4->5,
  1->3->4,
  2->6
]
merging them into one sorted list:
1->1->2->3->4->4->5->6

Example 2:

Input: lists = []
Output: []

Example 3:

Input: lists = [[]]
Output: []

Constraints:

• k == lists.length
• 0 <= k <= 10^4
• 0 <= lists[i].length <= 500
• -10^4 <= lists[i][j] <= 10^4
• lists[i] is sorted in ascending order.
• The sum of lists[i].length won't exceed 10^4.

---

小跟堆解题，放入每个有序链表的表头，小跟堆poll出最小的的Node，串联在输出的Node上。

public static class ListNodeComparator implements Comparator {

        @Override
        public int compare(ListNode o1, ListNode o2) {
            return o1.val - o2.val;
        }

    }

    public static ListNode mergeKLists(ListNode[] lists) {
        if (lists == null) {
            return null;
        }
        PriorityQueue heap = new PriorityQueue<>(new ListNodeComparator());
        for (int i = 0; i < lists.length; i++) {
            if (lists[i] != null) {
                heap.add(lists[i]);
            }
        }
        if (heap.isEmpty()) {
            return null;
        }
        ListNode head = heap.poll();
        ListNode pre = head;
        if (pre.next != null) {
            heap.add(pre.next);
        }
        while (!heap.isEmpty()) {
            ListNode cur = heap.poll();
            pre.next = cur;
            pre = cur;
            if (cur.next != null) {
                heap.add(cur.next);
            }
        }
        return head;
    }

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