Merge k Sorted Lists(合并k个有序链表)leetcode23

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

code1：（暴力破解）

public ListNode mergeKLists(ListNode[] lists) {
        if(lists==null||lists.length<=0)
            return null;
        if(lists.length==1)
            return lists[0];
        ListNode head=new ListNode(-1);
        ListNode result=head;
        int empty=0;
        int j;
            while(empty!=lists.length){
                empty=0;
                int minV=Integer.MAX_VALUE;
                int minI=-1;
                for(j=0;j<lists.length;j++){                 
                ListNode node=lists[j];
                    if(node==null){
                        empty++;
                        continue;
                        
                    }
                    if(node.val<minV){
                        minV=node.val;
                        minI=j;
                    }       
                }
                if(minI!=-1){
                    //System.out.println(minI);
                    head.next=lists[minI];
                    head=head.next;
                    lists[minI]=lists[minI].next;
                } 
            }
        return result.next;
        
    }

code2：（大优化）

 public ListNode mergeKLists(ListNode[] lists) {
        if(lists.length==0)
            return null;
        return merge(0,lists.length-1,lists);
    }
    public ListNode merge(int i,int j,ListNode[] lists) {
        if(j<i)return null;
        if(i==j)return lists[i];
        int mid=i+(j-i)/2;
        ListNode l=merge(i,mid,lists);
        ListNode r= merge(mid+1,j,lists);
        ListNode dummy =new ListNode(0);
        ListNode runner= dummy;

        while(l!=null && r!=null) {
            if(l.val>r.val) {
                runner.next=r;
                r=r.next;
                runner=runner.next;
            }
            else {
                runner.next=l;
                l=l.next;
                runner=runner.next;
            }
        }
        if(l==null && r==null)
            return dummy.next;
        if(l==null)
            runner.next=r;
        else
            runner.next=l;
        return dummy.next;
    }