【Day 87 】2023-09-04 - 23.合并 K 个排序链表

[azl397985856]

23.合并 K 个排序链表

入选理由

暂无

题目地址

https://leetcode-cn.com/problems/merge-k-sorted-lists/

前置知识

• 堆
• 链表
• 分而治之

  题目描述

  
  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

[Diana21170648]

思路

归并排序

代码

#先合并两个，两两合并最后再合到一起
class Solution:
    def mergeKLists(self, lists: List[ListNode]) -> ListNode:
        n = len(lists)

        # basic cases
        if n == 0: return None
        if n == 1: return lists[0]
        if n == 2: return self.mergeTwoLists(lists[0], lists[1])

        # divide and conqure if not basic cases
        mid = n // 2
        return self.mergeTwoLists(self.mergeKLists(lists[:mid]), self.mergeKLists(lists[mid:n]))

    def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
        res = ListNode(0)
        c1, c2, c3 = l1, l2, res
        while c1 or c2:
            if c1 and c2:
                if c1.val < c2.val:
                    c3.next = ListNode(c1.val)
                    c1 = c1.next
                else:
                    c3.next = ListNode(c2.val)
                    c2 = c2.next
                c3 = c3.next
            elif c1:
                c3.next = c1
                break
            else:
                c3.next = c2
                break

        return res.next

复杂度分析

• 时间复杂度：O(Nlogk)，其中 N 为链表所有节点的个数，logk为归并排序的层数。
• 空间复杂度：O(logk)

[Fuku-L]

代码

/**
 * 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 {
    class Status implements Comparable<Status>{
        int val;
        ListNode ptr;
        Status(int val, ListNode ptr){
            this.val = val;
            this.ptr = ptr;
        }
        public int compareTo(Status status2){
            return this.val - status2.val;
        }
    }
    PriorityQueue<Status> queue = new PriorityQueue<Status>();

    public ListNode mergeKLists(ListNode[] lists) {
        for(ListNode node: lists){
            if(node != null){
                queue.offer(new Status(node.val, node));
            }
        }
        ListNode head = new ListNode(0);
        ListNode tail = head;
        while(!queue.isEmpty()){
            Status f = queue.poll();
            tail.next = f.ptr;
            tail = tail.next;
            if(f.ptr.next != null){
                queue.offer(new Status(f.ptr.next.val, f.ptr.next));
            }
        }
        return head.next;
    }
}

[Alexno1no2]

class Solution:
    def mergeKLists(self, lists: List[Optional[ListNode]]) -> Optional[ListNode]:
        def mergeTwoLists(l1, l2):
            dummy = ListNode()
            current = dummy

            while l1 and l2:
                if l1.val < l2.val:
                    current.next = l1
                    l1 = l1.next
                else:
                    current.next = l2
                    l2 = l2.next
                current = current.next

            if l1:
                current.next = l1
            if l2:
                current.next = l2

            return dummy.next

        def mergeSortLists(lists, left, right):
            if left == right:
                return lists[left]
            mid = left + (right - left) // 2
            left_list = mergeSortLists(lists, left, mid)
            right_list = mergeSortLists(lists, mid + 1, right)
            return mergeTwoLists(left_list, right_list)

        if not lists:
            return None

        return mergeSortLists(lists, 0, len(lists) - 1)