Description:
You are given two non-empty linked lists representing two non-negative integers. 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.
You may assume the two numbers do not contain any leading zero, except the number 0 itself.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
算法思想：創建鏈表，逐位相加，逢十進一，保存進位，加到下一位中運算
C++代碼如下：
#include <iostream>
#include "stdlib.h"
using namespace std;
struct ListNode {
int val;
ListNode *next;
ListNode(int x) : val(x), next(NULL) {}
};
class Solution {
public:
ListNode* addTwoNumbers(ListNode *l1, ListNode *l2) {
int v1,v2,j=0,v;
ListNode *head,*p1,*p2;
head=p1=NULL;
while(l1!=NULL||l2!=NULL){
if(l1!=NULL){
v1=l1->val;
l1=l1->next;
}else{
v1=0;
}
if(l2!=NULL){
v2=l2->val;
l2=l2->next;
}else{
v2=0;
}
if(v1+v2+j<10){
v=v1+v2+j;
j=0;
}else{
v=v1+v2+j-10;
j=1;
}
p2=(ListNode *)malloc(sizeof(ListNode));
p2->val=v;
p2->next=NULL;
if(head==NULL){
head=p1=p2;
}else{
p1->next=p2;
p1=p1->next;
}
}
if(j==1){
p2=(ListNode *)malloc(sizeof(ListNode));
p2->val=1;
p2->next=NULL;
if(head==NULL){
head=p1=p2;
}else{
p1->next=p2;
p1=p1->next;
}
}
return head;
}
};
int main(){
ListNode *l1h,*p1,*p2,*l2h,*q1,*q2,*res;
p1=NULL;
l1h=NULL;
q1=NULL;
l2h=NULL;
int n,t,m,s;
cout<<"please input the first List element number:";
cin>>n;
for(int i=0;i<n;i++){
p2=(ListNode *)malloc(sizeof(ListNode));
cin>>t;
p2->val=t;
p2->next=NULL;
if(l1h==NULL){
l1h=p1=p2;
}else{
p1->next=p2;
p1=p1->next;
}
}
cout<<"please input the second List element number:";
cin>>m;
for(int i=0;i<m;i++){
q2=(ListNode *)malloc(sizeof(ListNode));
cin>>s;
q2->val=s;
q2->next=NULL;
if(l2h==NULL){
l2h=q1=q2;
}else{
q1->next=q2;
q1=q1->next;
}
}
// while(l1h!=NULL){
// cout<<l1h->val<<" ";
// l1h=l1h->next;
// }
// cout<<endl;
// while(l2h!=NULL){
// cout<<l2h->val<<" ";
// l2h=l2h->next;
// }
Solution sol;
res=sol.addTwoNumbers(l1h,l2h);
while(res!=NULL){
cout<<res->val<<" ";
res=res->next;
}
}
