[LeetCode] Merge Two Sorted Lists, Solution

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.
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[Thoughts]
简单的实现，也没什么可说的。

[Code]

1:    ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {  
2:      if(l1 == NULL) return l2;  
3:      if(l2 == NULL) return l1;  
4:      ListNode *head = new ListNode(-1);  
5:      ListNode *p = head;  
6:      while(l1 != NULL && l2!=NULL)  
7:      {  
8:          if(l1->val < l2->val)  
9:          {  
10:             p->next = l1;  
11:             l1= l1->next;  
12:          }  
13:          else  
14:          {  
15:             p->next = l2;  
16:             l2 = l2->next;  
17:          }  
18:          p = p->next;  
19:      }  
20:      if(l1 != NULL)  
21:          p->next = l1;  
22:      if(l2 != NULL)  
23:          p->next = l2;  
24:      p = head->next;  
25:      delete head;  
26:      return p;      
27:    }  

Update 04/13/13 refactor code for succinct

1:       ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {  
2:            ListNode* head = new ListNode(-1);  
3:            ListNode* p = head;  
4:            while(l1!=NULL || l2!= NULL)  
5:            {  
6:                 int val1 = l1==NULL?INT_MAX:l1->val;  
7:                 int val2 = l2==NULL? INT_MAX:l2->val;  
8:                 if(val1<=val2)  
9:                 {  
10:                      p->next = l1;          
11:                      l1=l1->next;  
12:                 }  
13:                 else  
14:                 {  
15:                      p->next = l2;  
16:                      l2 = l2->next;  
17:                 }  
18:                 p= p->next;  
19:            }  
20:            p = head->next;  
21:            delete head;  
22:            return p;  
23:       }