[LeetCode]Merge k Sorted Lists

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

思路：我的第一个想法是将lists中的链表两两合并排序，这样时间复杂度是o（n），n为所有链表中的数据，结果Time Limit Exceeded了。。。暂时没有想到太好的其他方法，希望有大神能给点思路，我先抛个砖，以下是我未AC的代码：

 

 1 /**
 2  * Definition for singly-linked list.
 3  * struct ListNode {
 4  *     int val;
 5  *     ListNode *next;
 6  *     ListNode(int x) : val(x), next(NULL) {}
 7  * };
 8  */
 9 class Solution {
10 public:
11     ListNode *mergeKLists(vector<ListNode *> &lists) {
12         if(lists.size() == 0)   return NULL;
13         
14         ListNode *pList=lists[0];
15         
16         vector<ListNode *>::iterator it=lists.begin()+1;
17         for(;it!=lists.end();++it)
18         {
19             pList=MergeLists(pList,*it);
20         }
21         
22         //for(int i=1;i<lists.size();i++)
23         //{
24             //pList=MergeLists(pList,lists[i]);
25         //}
26         
27         return pList;
28     }
29     
30     ListNode *MergeLists(ListNode *list1, ListNode *list2)
31     {
32         ListNode *pList=new ListNode(0);
33         ListNode *pHead=pList;
34         
35         while(list1 && list2)
36         {
37             if(list1->val > list2->val)
38             {
39                 pList->next=list2;
40                 list2=list2->next;
41             }
42             else
43             {
44                 pList->next=list1;
45                 list1=list1->next;
46             }
47             pList=pList->next;
48         }
49         
50         if(!list1)
51         {
52             pList->next=list2;
53         }
54         if(!list2)
55         {
56             pList->next=list1;
57         }
58         
59         pList=pHead;
60         pList=pList->next;
61         pHead->next=NULL;
62         delete(pHead);
63         
64         return pList;
65     }
66 };
View Code

 思路2：今天早上想到用归并排序的思想做一做，时间复杂度可以优化到o(nlogn),感觉很不错啊！于是说写就写，一次AC！好兴奋！

 1 /**
 2  * Definition for singly-linked list.
 3  * struct ListNode {
 4  *     int val;
 5  *     ListNode *next;
 6  *     ListNode(int x) : val(x), next(NULL) {}
 7  * };
 8  */
 9 class Solution {
10 public:
11     ListNode *mergeKLists(vector<ListNode *> &lists) {
12         if(lists.size() == 0)   return NULL;
13         
14         if(lists.size() == 1)   return lists[0];
15         
16         ListNode *pList=mergeKLists(lists, 0, lists.size()-1);
17         
18         return pList;
19     }
20     
21     ListNode *mergeKLists(vector<ListNode *> &lists, int left, int right)
22     {
23         if(left<right)
24         {
25             int mid=left+((right-left)>>1);
26             ListNode *pLeft=mergeKLists(lists,left,mid);
27             ListNode *pRight=mergeKLists(lists,mid+1,right);
28             return MergeLists(pLeft,pRight);
29         }
30         return lists[left];
31     }
32     
33     ListNode *MergeLists(ListNode *list1, ListNode *list2)
34     {
35         ListNode *pList=new ListNode(0);
36         ListNode *pHead=pList;
37         
38         while(list1 && list2)
39         {
40             if(list1->val > list2->val)
41             {
42                 pList->next=list2;
43                 list2=list2->next;
44             }
45             else
46             {
47                 pList->next=list1;
48                 list1=list1->next;
49             }
50             pList=pList->next;
51         }
52         
53         if(!list1)
54         {
55             pList->next=list2;
56         }
57         if(!list2)
58         {
59             pList->next=list1;
60         }
61         
62         pList=pHead;
63         pList=pList->next;
64         pHead->next=NULL;
65         delete(pHead);
66         
67         return pList;
68     }
69 };

 

本博客内容与代码均为作者Jarvis原创，如若转载请注明。

posted @ 2014-11-16 17:45  Jarvis_Wu  阅读(240)  评论(0)  编辑  收藏  举报