Leetcode117: Add Two Numbers

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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) {        int flag=0;        ListNode* p = new ListNode(0);        ListNode* tmp = p;        while(l1 || l2)        {            int val1 = 0;            if(l1)            {                val1 = l1->val;                l1 = l1->next;            }                        int val2 = 0;            if(l2)            {               val2 = l2->val;               l2 = l2->next;            }                        int sum = val1 + val2 + flag;            tmp->next = new ListNode(sum%10);            flag = sum/10;            tmp = tmp->next;        }        if(flag == 1)            tmp->next = new ListNode(1);                return p->next;    }};

python代码：

# Definition for singly-linked list.# class ListNode(object):#     def __init__(self, x):#         self.val = x#         self.next = Noneclass Solution(object):    def addTwoNumbers(self, l1, l2):        """        :type l1: ListNode        :type l2: ListNode        :rtype: ListNode        """       l3 = ListNode(-1)       head = l3       flag = 0       sum = 0       while l1!=None and l2!=None:       sum = l1.val + l2.val + flag       flag = sum/10       sum %= 10       head.next = ListNode(sum)       head = head.next       l1 = l1.next       l2 = l2.next       while l1!=None:       sum = l1.val + flag       flag = sum / 10       sum %= 10       head.next = ListNode(sum)       head = head.next       l1 = l1.next       while l2!=None:       sum = l2.val + flag        flag = sum / 10        sum %= 10        head.next = new ListNode(sum)        head = head.next        l2 = l2.next        if flag:        head.next = ListNode(flag)        return l3.next

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