Merge k Sorted Lists

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

 

第一遍：

 1 public class Solution {
 2     public ListNode mergeKLists(ArrayList<ListNode> lists) {
 3         // Start typing your Java solution below
 4         // DO NOT write main() function
 5         if(lists.size()==0) return null;
 6         ListNode header = new ListNode(-1), cur = header;
 7         // boolean sig = true;
 8         while(true){
 9             int pos = -1;
10             for(int i = 0; i < lists.size(); i ++){
11                 if(lists.get(i) != null && (pos == -1 || lists.get(i).val < lists.get(pos).val))
12                     pos = i;
13             }
14             if(pos != -1){
15                 cur.next = lists.get(pos);
16                 cur = cur.next;
17                 lists.set(pos,lists.get(pos).next);
18 
19             }else{
20                 break;
21             }
22         }
23         return header.next;
24     }
25 }

 

第三遍：

去掉 null的项 可以加速 查找。

 1 public class Solution {
 2     public ListNode mergeKLists(ArrayList<ListNode> lists) {
 3         ListNode header = new ListNode(-1);
 4         ListNode cur = header;
 5         if(lists == null || lists.size() == 0) return null;
 6         while(true){
 7             int min = -1;
 8             for(int i = 0; i < lists.size(); i ++){
 9                 if(lists.get(i) == null){
10                     lists.remove(i);
11                     i --;
12                 }
13                 else if(min == -1 || lists.get(i).val < lists.get(min).val)
14                     min = i;
15             }
16             if(min == -1) break;
17             else{
18                 cur.next = lists.get(min);
19                 lists.set(min,lists.get(min).next);
20                 cur = cur.next;
21             }
22         }
23         return header.next;
24     }
25 }

 

posted on 2014-08-06 10:29  Step-BY-Step  阅读(141)  评论(0)  编辑  收藏  举报

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