LeetCode刷题之第一题——Add Two Numbers

You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)

Output: 7 -> 0 -> 8

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
ListNode* p1=l1,*p2=l2;
ListNode* result=NULL,*temp=NULL,*presult=NULL;//result为结果链表头，temp为每位上相加后的值，presult为将temp链接到result上
int up=0;
while(p1!=NULL&&p2!=NULL){
temp= new ListNode(p1->val+p2->val);
temp->val=up+temp->val;//先加上进位得到本位之和
up=temp->val/10; //得到本位之和的进位值
temp->val=temp->val%10;//得到本位值
if(result==NULL){//空链表第一次赋值
result=temp;
presult=result;
}
else{
presult->next=temp;
presult=temp;
}
p1=p1->next;
p2=p2->next;

}
while(p2!=NULL){
temp= new ListNode(p2->val);
temp->val=up+temp->val;
up=temp->val/10;
temp->val=temp->val%10;
if(result==NULL){
result=temp;
presult=result;
}
else{
presult->next=temp;
presult=temp;
}
p2=p2->next;
}
while(p1!=NULL){
temp= new ListNode(p1->val);
temp->val=up+temp->val;
up=temp->val/10;
temp->val=temp->val%10;
if(result==NULL){
result=temp;
presult=result;
}
else{
presult->next=temp;
presult=temp;
}
p1=p1->next;
}
if(up>0)//[5] [5]结果为[0][1]而不是[0]
{
temp= new ListNode(up);
presult->next=temp;
}
return result;
}
};