23. Merge k Sorted Lists

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

最简单的思路是利用之前Merge 2 sorted Lists, 逐个相加，但是这种的时间复杂度会更高。

public ListNode MergeKLists(ListNode[] lists) {
        ListNode sentinel = null;
        for(int i = 0;i<lists.Count();i++)
        {
            sentinel = MergeTwoLists(sentinel,lists[i]);
        }
        return sentinel;
    }
    
    public ListNode MergeTwoLists(ListNode l1, ListNode l2) {
       if(l1 == null) return l2;
       if(l2 == null) return l1;
       var w1 = l1;
       var w2 = l2;
       var sentinel = new ListNode(-1);
       var dummy = sentinel;
       while(l1 != null && l2 != null)
       {
           if(l1.val >= l2.val) 
           {
               sentinel.next = new ListNode(l2.val);
               sentinel = sentinel.next;
               l2 = l2.next;
           }
           else
           {
               sentinel.next = new ListNode(l1.val);
               sentinel = sentinel.next;
               l1 = l1.next;
           }
       }
       if(l1 == null)
       sentinel.next = l2;
       else
       sentinel.next = l1;
       return dummy.next;
    }

下面用跟merge sort一样的思路，merge 为k/2的linked list array, 然后merge

public ListNode MergeKLists(ListNode[] lists) {
        var size = lists.Count();
        if(size==0) return null;
        if(size==1) return lists[0];
        while(size>1)
        {
            int k = (size+1)/2;
            for(int i =0;i<size/2;i++)
            {
                lists[i] = MergeTwoLists(lists[i], lists[i+k]);
            }
            size = k;
        }
        return lists[0];
    }
    
    public ListNode MergeTwoLists(ListNode l1, ListNode l2) {
       if(l1 == null) return l2;
       if(l2 == null) return l1;
       var w1 = l1;
       var w2 = l2;
       var sentinel = new ListNode(-1);
       var dummy = sentinel;
       while(l1 != null && l2 != null)
       {
           if(l1.val >= l2.val) 
           {
               sentinel.next = new ListNode(l2.val);
               sentinel = sentinel.next;
               l2 = l2.next;
           }
           else
           {
               sentinel.next = new ListNode(l1.val);
               sentinel = sentinel.next;
               l1 = l1.next;
           }
       }
       if(l1 == null)
       sentinel.next = l2;
       else
       sentinel.next = l1;
       return dummy.next;
    }