Leetcode 2. Add Two Numbers

Leetcode 2. Add Two Numbers

You are given two non-empty linked lists representing two non-negative integers. The digits are stored inreverse 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.

Example：

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)Output: 7 -> 0 -> 8Explanation: 342 + 465 = 807.

C++

/** * 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 carry = 0;        ListNode* listNode = new ListNode(0);    //创建并初始化listNode        ListNode* p1 = l1, *p2 = l2, *p3 = listNode;                 while(p1 != NULL | p2 != NULL)   //两链表不为空        {            if(p1 != NULL)      //l1链表不为空            {                carry += p1->val;  //                p1 = p1->next;            }            if(p2 != NULL)            {                carry += p2->val;                p2 = p2->next;            }            p3->next = new ListNode(carry%10);   // 存入p3链表中            p3 = p3->next;            carry /= 10;        }        if(carry == 1)        {            p3->next = new ListNode(1);  //存入p3链表中        }        return listNode->next;    }};

Python：

# Definition for singly-linked list.# class ListNode:#     def __init__(self, x):#         self.val = x#         self.next = Noneclass Solution:    def addTwoNumbers(self, l1, l2):        """        :type l1: ListNode        :type l2: ListNode        :rtype: ListNode        """        def V(val):            return val is not None        result = None        pre = None        tag = 0        while V(l1) or V(l2):            l1_val = V(l1) and l1.val or 0            l2_val = V(l2) and l2.val or 0            val = l1_val + l2_val + tag            tag =val / 10              # /10得出进位            node = ListNode(int(val % 10))  # ％10得出当前位的值            if result is None:                result = node                pre = node            else:                pre.next = node                pre =node            if V(l1):                l1 = l1.next            if V(l2):                l2 = l2.next                if tag == 1:    # 判断一次进位的值,需要补上一个node(多出一位)            pre.next = ListNode(1)           return result

参考：

http://blog.csdn.net/nomasp/article/details/48413977

http://blog.csdn.net/amds123/article/details/68942726  -Python