*Merge k Sorted Lists

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

leetcode上返回类型不太一样，改编一下第三个解法：4ms

public class Solution {
    public ListNode mergeKLists(ListNode[] lists) 
    {
            if (lists == null || lists.length == 0) {
            return null;
        }
        
        ListNode[] newlists = new ListNode[0];
        while(lists.length>1)
        {
           if(lists.length%2==0)
           {
               newlists = new ListNode[lists.length/2];
               for(int i=0;i<lists.length-1;i += 2)
               {
                    ListNode node = mergeTwoLists(lists[i],lists[i+1]);
                    newlists[i/2] = node;
               }
           }
           else
           {
               newlists = new ListNode[lists.length/2+1];
               for(int i=0;i<lists.length-1;i += 2)
               {
                    ListNode node = mergeTwoLists(lists[i],lists[i+1]);
                    newlists[i/2] = node;
               }
               newlists[newlists.length-1] = lists[lists.length-1];
           }
           lists = newlists;
            
        }
        return lists[0];
        
    }
    
    
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) 
    {
        ListNode dummyhead = new ListNode(0);
        ListNode head = dummyhead;
        while(l1!=null&&l2!=null)
        {
            if(l1.val<l2.val)
            {
                head.next = l1;
                l1 = l1.next;
            }
            else
            {
                head.next = l2;
                l2 = l2.next;
            }
            head = head.next;
        }
        
        if(l1!=null) head.next = l1;
        if(l2!=null) head.next = l2;
        return dummyhead.next;
        
    }
}

解法二：用heap, 12ms, 时间复杂度O(nlog(k))

Suppose the total number of nodes is n The total time complexity if (n * log k) .For a priority queue, insertion takes logK time

/** 
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    private Comparator<ListNode> ListNodeComparator = new Comparator<ListNode>() {
        public int compare(ListNode left, ListNode right) {
            if (left == null) {
                return 1;
            } else if (right == null) {
                return -1;
            }
            return left.val - right.val;
        }
    };
    
    public ListNode mergeKLists(ListNode[] lists) 
    {
        if(lists==null||lists.length==0)return null;
        Queue<ListNode> heap = new PriorityQueue<ListNode>(lists.length,ListNodeComparator);
        for(int i=0;i<lists.length;i++)
        {
            if(lists[i]!=null)
            {
                heap.add(lists[i]);
            }
        }
        ListNode dummy = new ListNode(0);
        ListNode tail = dummy;
        while(!heap.isEmpty())
        {
            ListNode head = heap.poll();
            tail.next = head;
            tail = head;
            if (head.next != null) {
                heap.add(head.next);
            }
        }
        return dummy.next;
    }
}