LeetCode - 21. Merge two sorted lists

Description

Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.

Difficulty

Solution

Idea

由于两个列表已排好序，因此可以设置两个下标i，j，分别指向l1和l2的首节点，再从头到尾分别比较两个列表当前下标对应的元素的大小关系，接着存到新列表（注意，这里题干要求“return it as a new list”）。需要留意两个列表的长度不一致的情况。

python

 1 # Definition for singly-linked list.
 2 # class ListNode(object):
 3 #     def __init__(self, x):
 4 #         self.val = x
 5 #         self.next = None
 6 
 7 class Solution(object):
 8     def mergeTwoLists(self, l1, l2):
 9         """
10         :type l1: ListNode
11         :type l2: ListNode
12         :rtype: ListNode
13         """
14         if l1 == None and l2 == None:
15             return None
16 
17         result = ListNode(-1)
18         result_cur = result
19 
20         ptr1 = l1
21         ptr2 = l2
22 
23         while ptr1 and ptr2:
24             if ptr1.val <= ptr2.val:
25                 new_node = ListNode(ptr1.val)
26                 result_cur.next = new_node
27                 ptr1 = ptr1.next
28             else:
29                 new_node = ListNode(ptr2.val)
30                 result_cur.next = new_node
31                 ptr2 = ptr2.next
32 
33             result_cur = result_cur.next
34 
35         while ptr1:
36             new_node = ListNode(ptr1.val)
37             result_cur.next = new_node
38             result_cur = result_cur.next
39             ptr1 = ptr1.next
40 
41         while ptr2:
42             new_node = ListNode(ptr2.val)
43             result_cur.next = new_node
44             result_cur = result_cur.next
45             ptr2 = ptr2.next
46 
47         return result.next
View Code

 

cpp

 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* mergeTwoLists(ListNode* l1, ListNode* l2) {
12         if (l1==NULL && l2==NULL)
13         {
14             return NULL;
15         }
16         ListNode* ptr1 = l1;
17         ListNode* ptr2 = l2;
18         ListNode* result = new ListNode(-1);
19         ListNode* result_cur = result;
20 
21         while (ptr1 && ptr2)
22         {
23             if (ptr1->val <= ptr2->val)
24             {
25                 result_cur->next = new ListNode(ptr1->val);;
26                 ptr1 = ptr1->next;
27             }else
28             {
29                 result_cur->next = new ListNode(ptr2->val);
30                 ptr2 = ptr2->next;
31             }
32             result_cur = result_cur->next;
33         }
34         while (ptr1)
35         {
36             result_cur->next = new ListNode(ptr1->val);
37             result_cur = result_cur->next;
38             ptr1 = ptr1->next;
39         }
40         while (ptr2)
41         {
42             result_cur->next = new ListNode(ptr2->val);;
43             result_cur = result_cur->next;
44             ptr2 = ptr2->next;
45         }
46 
47         return result->next;
48     }
49 };
View Code

 

 
posted @ 2016-09-15 19:48  朝研行歌  阅读(147)  评论(0)  编辑  收藏  举报