【Day 69 】2023-08-17 - 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

• 链表
• 归并排序

  题目描述

合并  k  个排序链表，返回合并后的排序链表。请分析和描述算法的复杂度。

示例:

输入: [   1->4->5,   1->3->4,   2->6 ] 输出: 1->1->2->3->4->4->5->6

[Beanza]

function mergeTwoLists(l1, l2) { const dummyHead = {}; let current = dummyHead; // l1: 1 -> 3 -> 5 // l2: 2 -> 4 -> 6 while (l1 !== null && l2 !== null) { if (l1.val < l2.val) { current.next = l1; // 把小的添加到结果链表 current = current.next; // 移动结果链表的指针 l1 = l1.next; // 移动小的那个链表的指针 } else { current.next = l2; current = current.next; l2 = l2.next; } }

if (l1 === null) { current.next = l2; } else { current.next = l1; } return dummyHead.next; }

[Diana21170648]

思路

先合并两个，两两合并最后再合到一起

代码

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(knlogk)，其中 N 为数组长度。
• 空间复杂度：O(logk)

[GuitarYs]

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

def mergeKLists(lists):
    if not lists:
        return None

    while len(lists) > 1:
        merged_lists = []
        for i in range(0, len(lists), 2):
            l1 = lists[i]
            l2 = lists[i+1] if i+1 < len(lists) else None
            merged = mergeTwoLists(l1, l2)
            merged_lists.append(merged)
        lists = merged_lists

    return lists[0]

def mergeTwoLists(l1, l2):
    if l1 is None:
        return l2
    if l2 is None:
        return l1

    if l1.val < l2.val:
        l1.next = mergeTwoLists(l1.next, l2)
        return l1
    else:
        l2.next = mergeTwoLists(l1, l2.next)
        return l2

[Fuku-L]

代码

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)