This problem can be solved using a priority queue in Python. Here are the steps:

1. Initialize a priority queue. In Python, you can use the `heapq` library which allows you to use a list as a heap.

2. For each linked list in the input, push the first element of the list along with the list itself into the priority queue. The priority queue will automatically sort the elements based on the first element of each list.

3. Pop the smallest element from the priority queue, and add it to the result list. Then, push the next element from the same list into the priority queue. If a list is empty after popping an element, just ignore it.

4. Repeat step 3 until the priority queue is empty.

5. Return the result list.

Here is the Python code for the above steps:

```python
import heapq

def mergeKLists(lists):
heap = [(lst[0], i, 0) for i, lst in enumerate(lists) if lst]
heapq.heapify(heap)
res = []
while heap:
val, list_ind, element_ind = heapq.heappop(heap)
res.append(val)
if element_ind + 1

Question

This problem can be solved using a priority queue in Python. Here are the steps:

1. Initialize a priority queue. In Python, you can use the `heapq` library which allows you to use a list as a heap.

2. For each linked list in the input, push the first element of the list along with the list itself into the priority queue. The priority queue will automatically sort the elements based on the first element of each list.

3. Pop the smallest element from the priority queue, and add it to the result list. Then, push the next element from the same list into the priority queue. If a list is empty after popping an element, just ignore it.

4. Repeat step 3 until the priority queue is empty.

5. Return the result list.

Here is the Python code for the above steps:

```python
import heapq

def mergeKLists(lists):
    heap = [(lst[0], i, 0) for i, lst in enumerate(lists) if lst]
    heapq.heapify(heap)
    res = []
    while heap:
        val, list_ind, element_ind = heapq.heappop(heap)
        res.append(val)
        if element_ind + 1 < len(lists[list_ind]):
            next_tuple = (lists[list_ind][element_ind + 1], list_ind, element_ind + 1)
            heapq.heappush(heap, next_tuple)
    return res
```

This code works by maintaining a heap of the smallest elements from each list. It repeatedly pops the smallest element from the heap and adds the next element from the same list into the heap, until all elements have been processed. The time complexity is O(N log k), where N is the total number of elements and k is the number of lists, because each insertion into a heap takes log k time and we do this for each of the N elements. The space complexity is O(k), for the heap.

Knowee AI · Accepted Answer