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.
归并排序两个链表
Solution:
采用归并排序思想即可,注意防止空指针的出现,在一个链表遍历完时需要把另一个没有遍历完的链表加在末尾。
Code:
class Solution {
public:
ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
ListNode* head=NULL;
if(!l1&&!l2)
return NULL;
else if(l1&&!l2)
return l1;
else if(!l1&&l2)
return l2;
else
{
if(l1->val<l2->val)
{
head=l1;
l1=l1->next;
}
else
{
head=l2;
l2=l2->next;
}
}//make sure which is head node
ListNode* temp=head;
while(l1&&l2)
{
if(l1->val<l2->val)
{
temp->next=l1;
temp=l1;
l1=l1->next;
}
else
{
temp->next=l2;
temp=l2;
l2=l2->next;
}
}
if(l1)//add rest nodes to the tail of link
temp->next=l1;
else
temp->next=l2;
return head;
}
};