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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 exceed104
.
- Time: O(n\log k)O(nlogk)
- Space: O(k)O(k)
Merge K Sorted List Program Solution In C++
class Solution {
public:
ListNode* mergeKLists(vector<ListNode*>& lists) {
ListNode dummy(0);
ListNode* curr = &dummy;
auto compare = [](ListNode* a, ListNode* b) { return a->val > b->val; };
priority_queue<ListNode*, vector<ListNode*>, decltype(compare)> minHeap(
compare);
for (ListNode* list : lists)
if (list)
minHeap.push(list);
while (!minHeap.empty()) {
ListNode* minNode = minHeap.top();
minHeap.pop();
if (minNode->next)
minHeap.push(minNode->next);
curr->next = minNode;
curr = curr->next;
}
return dummy.next;
}
};
Merge K Sorted List Program Solution In JAVA
class Solution {
public ListNode mergeKLists(ListNode[] lists) {
ListNode dummy = new ListNode(0);
ListNode curr = dummy;
Queue<ListNode> minHeap = new PriorityQueue<>((a, b) -> a.val - b.val);
for (final ListNode list : lists)
if (list != null)
minHeap.offer(list);
while (!minHeap.isEmpty()) {
ListNode minNode = minHeap.poll();
if (minNode.next != null)
minHeap.offer(minNode.next);
curr.next = minNode;
curr = curr.next;
}
return dummy.next;
}
}
Merge K Sorted List Program Solution In Python
from queue import PriorityQueue
class Solution:
def mergeKLists(self, lists: List[ListNode]) -> ListNode:
dummy = ListNode(0)
curr = dummy
pq = PriorityQueue()
for i, lst in enumerate(lists):
if lst:
pq.put((lst.val, i, lst))
while not pq.empty():
_, i, minNode = pq.get()
if minNode.next:
pq.put((minNode.next.val, i, minNode.next))
curr.next = minNode
curr = curr.next
return dummy.next