[LeetCode] Merge k Sorted Lists 解题报告

最新推荐文章于 2019-04-07 10:13:21 发布

codingtmd_in_china

最新推荐文章于 2019-04-07 10:13:21 发布

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本文链接：https://blog.youkuaiyun.com/codingtmd_in_china/article/details/50502572

本文介绍了一种使用归并排序思想解决合并K个已排序链表问题的方法，并提供了详细的C++实现代码。通过两两合并的方式逐步将所有链表整合成一个有序链表。

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

» Solve this problem

[解题思路]
merge sort。外面套一层循环即可。注意指针操作即可。

[Code]

1:    ListNode *mergeKLists(vector<ListNode *> &lists) {  
2:      // Start typing your C/C++ solution below  
3:      // DO NOT write int main() function  
4:      ListNode * head = new ListNode(INT_MIN);  
5:      for(int i = 0; i < lists.size(); i++)  
6:      {  
7:        ListNode* p1 = head->next;  
8:        ListNode * p2 = lists[i];  
9:        ListNode* pre = head;  
10:        while(p1!= NULL && p2!= NULL)  
11:        {  
12:          if(p1->val >= p2->val)  
13:          {  
14:            pre->next = p2;   
15:            p2 = p2->next; pre = pre->next;  
16:            pre->next = p1;  
17:            continue;  
18:          }  
19:          pre = p1;  
20:          p1 = p1->next;;  
21:        }  
22:        if(p2 != NULL)  
23:        {  
24:          pre->next = p2;  
25:        }  
26:      }  
27:      ListNode* del = head;  
28:      head = head->next;  
29:      delete del;  
30:      return head;      
31:    }

Update, 3/9/2013
Refactor the code

1:    ListNode *mergeKLists(vector<ListNode *> &lists) {  
2:      if(lists.size() == 0) return NULL;  
3:      ListNode *p = lists[0];  
4:      for(int i =1; i< lists.size(); i++)  
5:      {  
6:        p = merge2Lists(p, lists[i]);  
7:      }  
8:      return p;  
9:    }  
10:    ListNode * merge2Lists(ListNode *head1, ListNode *head2)  
11:    {  
12:      ListNode *head = new ListNode(INT_MIN);  
13:      ListNode *p = head;  
14:      while(head1!=NULL && head2!=NULL)  
15:      {  
16:        if(head1->val < head2->val)  
17:        {  
18:          p->next = head1;  
19:          head1 = head1->next;  
20:        }  
21:        else  
22:        {  
23:          p->next = head2;  
24:          head2 = head2->next;  
25:        }  
26:        p = p->next;  
27:      }  
28:      if(head1 !=NULL)  
29:      {  
30:        p->next = head1;  
31:      }  
32:      if(head2 != NULL)  
33:      {  
34:        p->next = head2;  
35:      }  
36:      p = head;  
37:      head = head->next;  
38:      delete p;  
39:      return head;  
40:    }