LintCode-两个链表的和sum

3->1->5->null

5->9->2->null

8->0->8->null.....

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
/**
* @param l1: the first list
* @param l2: the second list
* @return: the sum list of l1 and l2
*/
ListNode *addLists(ListNode *l1, ListNode *l2) {
// write your code here

ListNode *head;
ListNode *p1 = new ListNode;  //必须分配地址
int carry=0,carry1;
head=p1;
if (l1==NULL)
return l2;
if(l2==NULL)
return l1;
while(l1!=NULL && l2!=NULL){
p1->val=l1->val+l2->val+carry;
carry=(p1->val)/10;
carry1=(p1->val)%10;
if(carry==1){
p1->val=carry1;
}
l1=l1->next;
l2=l2->next;
if(l1!=NULL && l2!=NULL){
p1->next = new ListNode;  //必须分配地址
p1=p1->next;
}
}
if(l1==NULL){
while(l2!=NULL){
p1->next = new ListNode;  //dddddd
p1=p1->next;
p1->val=l2->val+carry;
carry=(p1->val)/10;
carry1=(p1->val)%10;
if(carry==1)
p1->val=carry1;
l2=l2->next;

}

}
if(l2==NULL){
while(l1!=NULL){
p1->next = new ListNode;  //ddddd
p1=p1->next;
p1->val=l1->val+carry;
carry=(p1->val)/10;
carry1=(p1->val)%10;
if(carry==1)
p1->val=carry1;
l1=l1->next;

}

}

if(carry==1){    //最后一个数看是不是大于10，
p1->next = new ListNode;
p1=p1->next;
p1->val=1;
p1->next=NULL;
}
else{
p1->next=NULL;
}

return head;

}
};