Binary Loop Solution - Merge k Sorted Lists **

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode mergeKLists(ArrayList<ListNode> lists) {
        if(lists.isEmpty()) return null;
        if(lists.size() == 1) return lists.get(0);
        // int k = lists.size();
        // int log = (int)(Math.log(k)/Math.log(2));
        // log = log < Math.log(k)/Math.log(2)? log+1:log; // take ceiling
        // for(int i = 1; i <= log; i++){
        //     for(int j = 0; j < lists.size(); j=j+(int)Math.pow(2,i)){
        //         int offset = j+(int)Math.pow(2,i-1);
        //         lists.set(j, mergeTwoLists(lists.get(j), (offset >= lists.size()? null : lists.get(offset))));
        //     }
        // }
        int index = lists.size();
        while(index > 1){
            int remain = index % 2;
            int middle = index / 2;
            for(int i = 0; i < middle; i++){
                lists.set(i + remain, mergeTwoLists(lists.get(i + remain), lists.get(i + middle + remain)));
            }
            index = middle + remain;
        }
        return lists.get(0);
    }
The idea is to binary merge the list heads in the list, and merge to first half. Notice for remaining case… Take the advantage that eventually the divide will be 1 by add the remaining to the next index each iteration!

    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        if(l1 == null) return l2;
        if(l2 == null) return l1;
        ListNode head = l1.val > l2.val? l2:l1;
        if(head.equals(l1)){
            l1 = l2;
            l2 = head;
        }
        while(l2.next != null && l1 != null){
            if(l2.next.val > l1.val){
                ListNode tmp = l2.next;
                l2.next = l1;
                l1 = l1.next;
                l2 = l2.next;
                l2.next = tmp;
            }
            else
                l2 = l2.next;
        }
        if(l1 != null){
            l2.next = l1;
        }
        return head;
    }
}