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
class Solution { public: ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) { if(l1==NULL && l2==NULL) return NULL; if (l1==NULL) return l2; if (l2==NULL) return l1; ListNode *head = NULL , *pre = NULL; int c=0; while( l1!=NULL || l2!=NULL || c!=0 ) { if(l1!=NULL) { c += l1->val; l1 = l1->next; } if(l2!=NULL) { c += l2->val; l2 = l2->next; } ListNode *p = new ListNode(c%10); if(head == NULL) head = p; else pre->next = p; pre = p; c /= 10; } return head; } };
#include<iostream> using namespace std; #define N 3 struct ListNode { int val; ListNode *next; ListNode(int x) : val(x), next(NULL) {} }; class Solution { public: ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) { if(l1==NULL && l2==NULL) return NULL; if (l1==NULL) return l2; if (l2==NULL) return l1; ListNode *head = NULL , *pre = NULL; int c=0; while( l1!=NULL || l2!=NULL || c!=0 ) { if(l1!=NULL) { c += l1->val; l1 = l1->next; } if(l2!=NULL) { c += l2->val; l2 = l2->next; } ListNode *p = new ListNode(c%10); if(head == NULL) head = p; else pre->next = p; pre = p; c /= 10; } return head; } }; ListNode *creatlist() { ListNode *head=NULL; for(int i=0; i<N; i++) { int a; cin>>a; ListNode *p; p = (ListNode*)malloc(sizeof(ListNode)); p->val = a; p->next = head; head = p; } return head; } int main() { ListNode *list1 = creatlist(); ListNode *list2 = creatlist(); Solution lin; ListNode *outlist = lin.addTwoNumbers( list1,list2 ); for(int i=0; i<N; i++) { cout<<outlist->val; outlist = outlist->next; } }
原文地址:http://blog.csdn.net/keyyuanxin/article/details/43817511