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

divide and conquer，用merge 2 sorted list作为函数，recursion

时间：O(nklogk)  　　-- n: average length of a linked list, merge: O(n)，空间：O(logk) 　　-- k: # of linked lists in lists

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        if(lists == null || lists.length == 0)
            return null;
        return mergeKLists(lists, 0, lists.length - 1);
    }
    
    private ListNode mergeKLists(ListNode[] lists, int l, int r) {
        while(l < r) {
            int m = l + (r - l) / 2;
            return mergeTwoLists(mergeKLists(lists, l, m), mergeKLists(lists, m + 1, r));
        }
        return lists[l];
    }
    
    private ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        ListNode dummy = new ListNode(-1);
        ListNode cur = dummy;
        while(l1 != null && l2 != null) {
            if(l1.val < l2.val) {
                cur.next = l1;
                l1 = l1.next;
            }
            else {
                cur.next = l2;
                l2 = l2.next;
            }
            cur = cur.next;
        }
        cur.next = l1 == null ? l2 : l1;
        
        return dummy.next;
    }
} 

ref: https://leetcode.com/problems/merge-k-sorted-lists/discuss/10610/C%2B%2B-code-O(NlogK)-in-time-O(1)-in-space-Divide_Conquer