[leetcode] 23. Merge k Sorted Lists

 

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 <= 104
• 0 <= lists[i].length <= 500
• -104 <= lists[i][j] <= 104
• lists[i] is sorted in ascending order.
• The sum of lists[i].length will not exceed 104.

Accepted

1.5M

Submissions

3M

Acceptance Rate

48.5%

# Definition for singly-linked list.
class ListNode(object):
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next
class Solution(object):
    def mergeKLists(self, lists):
        """
        :type lists: List[ListNode]
        :rtype: ListNode
        """
        root = result = ListNode(None)
        heap = [] 

        # 각 연결 리스트의 루트를 힙에 저장 
        for i in range(len(lists)):
            if lists[i]:
                heapq.heappush(heap, (lists[i].val, i, lists[i]))

        # 힙 추출 이후 다음 노드는 다시 저장 
        while heap: 
            node = heapq.heappop(heap)
            idx = node[1]
            result.next = node[2]

            result = result.next
            if result.next:
                heapq.heappush(heap, (result.next.val, idx, result.next))

        return root.next