Published: Sep 22, 2022
Like other linked list problems, start from creating a dummy node as a head. It’s a good idea to try a brute force approach if the problem is the linked list. Save values or nodes to an array, then apply necessary operation to the array. Make those values back to the linked list. If the tests’ memory/cpu consumptions are not so extensive, the brute force might be enough.
You are given an array of
lists, each linked-list is sorted in ascending order.
Merge all the linked-lists into one sorted linked-list and return it.
k == lists.length
0 <= k <= 10**4
0 <= lists[i].length <= 500
- -104 <= lists[i][j] <= 104`
lists[i]is sorted in ascending order.
- The sum of
lists[i].lengthwill not exceed 10**4.
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: 
Create a dummy node since we don’t know what linked list’s head becomes ground total linked list head. The cur pointer goes through all linked lists’ all nodes as if those are one. When all lists’ nodes are checked, the dummy node as a head has all linked lists nodes. Made the cur pointer back to dummy’s next. Update each list’s node value by sorted values’ each element.
# Definition for singly-linked list. # class ListNode: # def __init__(self, val=0, next=None): # self.val = val # self.next = next class MergeKSortedLists: def mergeKLists(self, lists: List[Optional[ListNode]]) -> Optional[ListNode]: dummy = ListNode() cur = dummy values =  for l in lists: while l: values.append(l.val) cur.next = l l = l.next cur = cur.next cur = dummy.next for v in sorted(values): cur.val = v cur = cur.next return dummy.next
O(n)– n: total number of list nodes