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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

思路:

sum=(s1+s2+carry)%10;
carry=(s1+s2+carry)/10;

由以上关系即可得到code;

code:

/** * 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 head(-1);        int carry=0;        ListNode *cur=&head;                for(ListNode *p1=l1,*p2=l2;p1!=NULL||p2!=NULL;            p1=p1==NULL ? NULL:p1->next,p2=p2==NULL ? NULL:p2->next,            cur=cur->next)        {            const int s1=p1==NULL ? 0:p1->val;            const int s2=p2==NULL ? 0:p2->val;            const int sum=(s1+s2+carry)%10;            carry=(s1+s2+carry)/10;            cur->next=new ListNode(sum);        }                if(carry>0)            cur->next=new ListNode(carry);                return head.next;    }};
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Add Two Numbers