【leetcode】Merge Two Sorted Lists

Merge Two Sorted Lists

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.

 

每次找到两个链表中小的一个，把他连接在后面，直到连接完成

 

思路1：利用递归方法解：递归思路：

       

        if(l1->val<=l2->val)

        {

            head=l1;

            head->next=mergeTwoLists(l1->next,l2);

        }

        else

        {

            head=l2;

            head->next=mergeTwoLists(l1,l2->next);

        }

 

 

 

 1 class Solution {

 2 public:

 3     ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {                  

 4          

 5         if(l1==NULL)

 6         {

 7             return l2;

 8         }

 9        

10         if(l2==NULL)

11         {

12             return l1;

13         }

14        

15         ListNode *head;

16        

17         if(l1->val<=l2->val)

18         {

19             head=l1;

20             head->next=mergeTwoLists(l1->next,l2);

21         }

22         else

23         {

24             head=l2;

25             head->next=mergeTwoLists(l1,l2->next);

26         }

27        

28         return head;

29        

30     }

31 };

 

 

 

 

非递归解法：

 

 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        

13         if(l1==NULL)

14         {

15             return l2;

16         }

17        

18         if(l2==NULL)

19         {

20             return l1;

21         }

22        

23         ListNode *head=new ListNode(0);

24         ListNode *pre=head;

25        

26         while(l1!=NULL||l2!=NULL)

27         {

28             if(l1==NULL)

29             {

30                 head->next=l2;

31                 l2=l2->next;

32                 head=head->next;

33                 continue;

34             }

35            

36             if(l2==NULL)

37             {

38                 head->next=l1;

39                 l1=l1->next;

40                 head=head->next;

41                 continue;

42             }

43            

44             if(l1->val<=l2->val)

45             {

46                 head->next=l1;

47                 l1=l1->next;

48                 head=head->next;

49             }

50             else

51             {

52                 head->next=l2;

53                 l2=l2->next;

54                 head=head->next;

55             }

56         }

57        

58         ListNode *ret=pre->next;

59         delete pre;

60         return pre->next;

61        

62     }

63 };