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 -> 8
Explanation: 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
4000
+= 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 = None

class 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