【算法】【线性表】【链表】合并 K 个升序链表

1  题目

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。

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

2  解答

底子是基于归并排序的思想，先拆再合，最后的落点其实就是两个有序链表的合并，跟两个有序数组的合并思想差不多哈：

/**
 * 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 == null || lists.length <= 0) {
            return null;
        }
        return mergeKListsWithSpilt(lists, 0, lists.length - 1);
    }

    private ListNode mergeKListsWithSpilt(ListNode[] lists, int start, int end) {
        if (start >= end) {
            return lists[start];
        }
        // 先拆分
        int middle = start + (end - start) / 2;
        ListNode leftNode = mergeKListsWithSpilt(lists, start, middle);
        ListNode rightNode = mergeKListsWithSpilt(lists, middle + 1, end);

        // 合并两个链表
        if (leftNode == null) {
            return rightNode;
        }
        if (rightNode == null) {
            return leftNode;
        }

        // 头节点
        ListNode head = new ListNode();
        ListNode res = head;
        while (leftNode != null && rightNode != null) {
            if (leftNode.val < rightNode.val) {
                head.next = leftNode;
                leftNode = leftNode.next;
            } else {
                head.next = rightNode;
                rightNode = rightNode.next;
            }
            head = head.next;
        }

        while (leftNode != null) {
            head.next = leftNode;
            head = head.next;
            leftNode = leftNode.next;
        }

        while (rightNode != null) {
            head.next = rightNode;
            head = head.next;
            rightNode = rightNode.next;
        }
        return res.next;
    }
}

加油。