Leetcode - Merge k Sorted Lists

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

[分析] 有了Merge 2 sorted lists的铺垫做这题就容易了，自己写了递归版的method1, 翻看历史记录，有迭代版的和heap sort版本，觉得method1 & 2写得更顺手些。感觉迭代方式挺巧妙的，哈哈

public class Solution {
    // Method 1: 递归
    public ListNode mergeKLists(ListNode[] lists) {
        if (lists == null || lists.length == 0)
            return null;
        return divide(lists, 0, lists.length - 1);
    }
    private ListNode divide(ListNode[] lists, int beg, int end) {
        if (beg != end) {
            int mid = beg + (end - beg) / 2;
            return merge2Lists(divide(lists, beg, mid), divide(lists, mid + 1, end));
        } else {
            return lists[beg];
        }
    }
    // Method 2: 迭代
    public ListNode mergeKLists2(ArrayList<ListNode> lists) {
        int last = lists.size() - 1;
        if(last < 0)
            return null;

        while(last > 0){
            int curr = 0;
            while(curr < last)
                lists.set(curr, merge2Lists(lists.get(curr++), lists.get(last--)));
        }
        return lists.get(0);
    }
    private ListNode merge2Lists(ListNode l1, ListNode l2) {
        if (l1 == null) return l2;
        if (l2 == null) return l1;
        ListNode head = new ListNode(0);
        ListNode m = head;
        ListNode p = l1, q = l2;
        while (p != null && q != null) {
            if (p.val < q.val) {
                m.next = p;
                p = p.next;
            } else {
                m.next = q;
                q = q.next;
            }
            m = m.next;
        }
        if (p != null)
            m.next = p;
        if (q != null)
            m.next = q;
        return head.next;
    }
    // Method 3: heap sort(O(nlogk)), n nodes, k lists
     public ListNode mergeKLists(ArrayList<ListNode> lists) {

        if(lists == null || lists.size() == 0)
            return null;

        Comparator<ListNode> comparator = new Comparator<ListNode>(){
            public int compare(ListNode l1, ListNode l2){
                if(l1.val < l2.val) 
                    return -1;
                else if(l1.val > l2.val) 
                    return 1;
                else 
                    return 0;
            }
        };

        //initial 
        PriorityQueue<ListNode> heap = new PriorityQueue<ListNode>(lists.size(), comparator);
        for(ListNode node:lists){
            if(node != null)
                heap.offer(node);
        }

        ListNode head = null, curr = head;
        while(!heap.isEmpty()){
            if(head == null){
                head = heap.poll();//delete min from heap
                curr = head;
            }else{
                curr.next = heap.poll();
                curr = curr.next;
            }
            if(curr.next != null)
                heap.offer(curr.next);
        }

        return head;    
    }
}