23. Merge k Sorted Lists

Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.

Example:

Input:
[
  1->4->5,
  1->3->4,
  2->6
]
Output: 1->1->2->3->4->4->5->6

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        int k = lists.length;
        ListNode dummy = new ListNode(0);
        ListNode []p = new ListNode[k];
        
        for (int i = 0; i < k; ++i) {
            p[i] = lists[i];
        }
        ListNode temp = dummy;
        while (true) {
            int pos = -1, min = -1;
            for (int i = 0; i < k; ++i) {
                if (p[i] != null) {
                    if (pos == -1 || p[i].val < min) {
                        pos = i;
                        min = p[i].val;
                    }
                }
            }
            if (pos == -1) break;
            
            temp.next = new ListNode(p[pos].val);
            temp = temp.next;
            p[pos] = p[pos].next;
            
        }
        return dummy.next;
        
    }
}

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */


class Solution {
    static Comparator<ListNode> cmp = new Comparator<ListNode>() {
        public int compare(ListNode e1, ListNode e2) {
            return e1.val - e2.val;
        }
    };
    public ListNode mergeKLists(ListNode[] lists) {
        int k = lists.length;
        ListNode dummy = new ListNode(0);
        ListNode p = dummy;
        Queue<ListNode> q = new PriorityQueue<>(cmp);
        
        
        for (int i = 0; i < k; ++i) {
            if (lists[i] != null) {
                q.add(lists[i]);
            }
        }
        
        while (!q.isEmpty()) {
            ListNode temp = q.poll();
            p.next = new ListNode(temp.val);
            if (temp.next != null) {
                q.add(temp.next);
            }
            p = p.next;
        }
        return dummy.next;
        
    }
}