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

英文原文

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.

中文题目

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。

 

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

通过代码

高赞题解

思路：

思路 1：

优先级队列

时间复杂度：$O(n*log(k))$，n 是所有链表中元素的总和，k 是链表个数。

思路 2：

分而治之

链表两两合并

代码：

思路 1：

[]
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def mergeKLists(self, lists: List[ListNode]) -> ListNode:
        import heapq
        dummy = ListNode(0)
        p = dummy
        head = []
        for i in range(len(lists)):
            if lists[i] :
                heapq.heappush(head, (lists[i].val, i))
                lists[i] = lists[i].next
        while head:
            val, idx = heapq.heappop(head)
            p.next = ListNode(val)
            p = p.next
            if lists[idx]:
                heapq.heappush(head, (lists[idx].val, idx))
                lists[idx] = lists[idx].next
        return dummy.next

[]
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
   public ListNode mergeKLists(ListNode[] lists) {
        if (lists == null || lists.length == 0) return null;
        PriorityQueue<ListNode> queue = new PriorityQueue<>(lists.length, new Comparator<ListNode>() {
            @Override
            public int compare(ListNode o1, ListNode o2) {
                if (o1.val < o2.val) return -1;
                else if (o1.val == o2.val) return 0;
                else return 1;
            }
        });
        ListNode dummy = new ListNode(0);
        ListNode p = dummy;
        for (ListNode node : lists) {
            if (node != null) queue.add(node);
        }
        while (!queue.isEmpty()) {
            p.next = queue.poll();
            p = p.next;
            if (p.next != null) queue.add(p.next);
        }
        return dummy.next;
    }
}

思路 2：

分而治之

[]
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def mergeKLists(self, lists: List[ListNode]) -> ListNode:
        if not lists:return 
        n = len(lists)
        return self.merge(lists, 0, n-1)
    def merge(self,lists, left, right):
        if left == right:
            return lists[left]
        mid = left + (right - left) // 2
        l1 = self.merge(lists, left, mid)
        l2 = self.merge(lists, mid+1, right)
        return self.mergeTwoLists(l1, l2)
    def mergeTwoLists(self,l1, l2):
        if not l1:return l2
        if not l2:return l1
        if l1.val < l2.val:
            l1.next = self.mergeTwoLists(l1.next, l2)
            return l1
        else:
            l2.next = self.mergeTwoLists(l1, l2.next)
            return l2

[]
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
   public ListNode mergeKLists(ListNode[] lists) {
        if (lists == null || lists.length == 0) return null;
        return merge(lists, 0, lists.length - 1);
    }

    private ListNode merge(ListNode[] lists, int left, int right) {
        if (left == right) return lists[left];
        int mid = left + (right - left) / 2;
        ListNode l1 = merge(lists, left, mid);
        ListNode l2 = merge(lists, mid + 1, right);
        return mergeTwoLists(l1, l2);
    }

    private ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if (l1 == null) return l2;
        if (l2 == null) return l1;
        if (l1.val < l2.val) {
            l1.next = mergeTwoLists(l1.next, l2);
            return l1;
        } else {
            l2.next = mergeTwoLists(l1,l2.next);
            return l2;
        }
    }
}

统计信息

通过次数	提交次数	AC比率
354917	632291	56.1%

提交历史

提交时间	提交结果	执行时间	内存消耗	语言

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