[LeetCode] 23. Merge k Sorted Lists

23. Merge k Sorted Lists

Hard

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.

---

문제 풀이

정렬된 링크드 리스트가 들어간 배열이 주어진다. 

여러개의 링크드 리스트를 가지고 하나의 정렬된 링크드 리스트를 만들어야한다.

모든 리스트의 노드들을 우선순위 큐, 즉 힙에 넣는다. 

힙에서 하나씩 빼어 답을 구성한다.

소스 코드

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def mergeKLists(self, lists: List[Optional[ListNode]]) -> Optional[ListNode]:
        ans = None
        total = []
        heapify(total)
        for l in lists:
            cur = l
            while cur:
                heappush(total, cur.val)
                cur = cur.next
        if len(total) == 0:
            return ans
        ans = ListNode(heappop(total), None)
        temp = ans
        while total:
            temp.next = ListNode(heappop(total), None)
            temp = temp.next
        return ans