leetcode--Add Two Numbers

来源：互联网发布：arch linux chroot 编辑：程序博客网时间：2024/06/04 19:53

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

题目简单，没什么值得优化的

AC代码:

/** * 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*result=NULL;        ListNode*lt1=l1;        ListNode*lt2=l2;        ListNode*tmp,*curr;        int carry=0;        int num;                result=(ListNode*)malloc(sizeof(ListNode));        result->next=NULL;        //result->val=9999;        curr=result;        while(lt1!=NULL&<2!=NULL){            if((num=lt1->val+lt2->val+carry)>9){                carry=1;                num%=10;            }else{                carry=0;            }            tmp=(ListNode*)malloc(sizeof(ListNode));            tmp->val=num;            tmp->next=NULL;            curr->next=tmp;            curr=tmp;                        lt1=lt1->next;            lt2=lt2->next;        }                if(lt1!=NULL){            while(lt1!=NULL){                                if((num=lt1->val+carry)>9){                    carry=1;                    num%=10;                }else{                    carry=0;                }                                tmp=(ListNode*)malloc(sizeof(ListNode));                tmp->val=num;                tmp->next=NULL;                curr->next=tmp;                curr=tmp;                lt1=lt1->next;            }        }        if(lt2!=NULL){            while(lt2!=NULL){                                 if((num=lt2->val+carry)>9){                    carry=1;                    num%=10;                }else{                    carry=0;                }                                tmp=(ListNode*)malloc(sizeof(ListNode));                tmp->val=num;                tmp->next=NULL;                curr->next=tmp;                curr=tmp;                lt2=lt2->next;            }        }                if(carry==1){            tmp=(ListNode*)malloc(sizeof(ListNode));            tmp->val=1;            tmp->next=NULL;            curr->next=tmp;            curr=tmp;        }        return result->next;    }};

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