Two approaches to solve the problem

• Approach 1:
We can create a priority queue which would store the heads of the lists which we want to merge and then just keep on fetching a node from the priority queue and adding it to the final list. Whenever we fetch the node from Priority Queue just add the next node of the list of whose node was added to final list.

``````/**
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) { val = x; }
* }
*/
public class Solution {
if(lists==null || lists.length==0) {
return null;
}

ListNode newNode = new ListNode(0);
ListNode temp = newNode;

PriorityQueue<ListNode> queue = new PriorityQueue<>(new ListComparator());

for(ListNode node : lists) {
if(node!=null) {
queue.offer(node);
}
}

while(!queue.isEmpty()) {
ListNode node = queue.poll();
temp.next = node;
temp = temp.next;

if(node.next!=null) {
queue.offer(node.next);
}
}

return newNode.next;
}

class ListComparator implements Comparator {
@Override
public int compare(Object o1,Object o2) {
ListNode l1 = (ListNode)o1;
ListNode l2 = (ListNode)o2;

return l1.val-l2.val;
}
}
}
``````

Approach 2:
Another approach to do solve this is to use the merge sort over the array of Lists. This is done by using the same merge sort algorithm like we do in an number array.

``````/**
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode mergeKLists(ListNode[] lists) {
if(lists==null || lists.length==0) {
return null;
}
return partition(lists,0,lists.length-1);
}
private ListNode partition(ListNode[] lists,int start,int end) {
if(start==end) {
return lists[start];
}
else if(start<end) {
int mid = start + (end-start)/2;
ListNode l1 = partition(lists,start,mid);
ListNode l2 = partition(lists,mid+1,end);
return merge(l1,l2);
}
else {
return null;
}
}
private ListNode merge(ListNode l1,ListNode l2) {
if(l1==null) {
return l2;
}
else if(l2==null) {
return l1;
}
else if(l1.val<l2.val) {
l1.next = merge(l1.next,l2);
return l1;
}
else {
l2.next = merge(l1,l2.next);
return l2;
}
}
}
``````

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