lc23. 合并K个升序链表(小根堆&递归&迭代）

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

/**
 * 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) {
        PriorityQueue<ListNode> q = new PriorityQueue<>((a,b) -> a.val-b.val);
        for (ListNode list : lists) {
            if (list != null) 
                q.add(list);
        }
        ListNode dummy = new ListNode(0);
        ListNode cur = dummy;
        while (!q.isEmpty()) {
            ListNode minNode = q.poll();
            cur.next = minNode;
            cur = cur.next;
            if (minNode.next != null) {
                q.add(minNode.next);
            }
        }
        return dummy.next;
    }
}

分治

/**
 * 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 || lists == null) return null;
        return merge(lists, 0, lists.length - 1);
    }
    private ListNode merge(ListNode[] lists, int l, int r) {
        if (l == r) return lists[l];
        int mid = l + r >> 1;
        ListNode l1 = merge(lists, l, mid);
        ListNode l2 = merge(lists, mid + 1, r);
        return mergeTwoLists(l1, l2);
    } 
    private ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if (l1 == null || l2 == null) 
            return l1 == null ? l2 : l1;
        if (l1.val <= l2.val) {
            l1.next = mergeTwoLists(l1.next, l2);
            return l1;
        } else {
            l2.next = mergeTwoLists(l2.next, l1);
            return l2;
        }
    }
}