题目：https://leetcode.com/problems/add-two-numbers/description/
题目描述
You are given two non-empty linked lists representing two non-negative integers. 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.
You may assume the two numbers do not contain any leading zero, except the number 0 itself.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4) Output: 7 -> 0 -> 8
方法
1.将两条list转置，方便计算获取两个加数。
2.将结果存入。
这个方法仅适用于python 考虑到很大的数，本题真正应该考察的是对运算的理解上，参见下面的 C++版。
python版：
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def addTwoNumbers(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
tmplist = None
a = 0;
b = 0;
while l1 :
tmpnode = l1
l1 = l1.next
tmpnode.next = tmplist;
tmplist = tmpnode;
while tmplist:
a = a*10+tmplist.val
tmplist = tmplist.next;
while l2:
tmpnode = l2
l2 = l2.next
tmpnode.next = tmplist;
tmplist = tmpnode;
while tmplist:
b = b*10+tmplist.val
tmplist = tmplist.next;
c = a + b
if c==0:
return ListNode(0)
tmplist = ListNode(0);
tmpnode = tmplist
while c!=0:
tmpnode.next = ListNode(c)
tmpnode = tmpnode.next
c/=10;
tmpnode = tmplist.next;
del tmplist;
return tmpnode;
C++版：
/**
* 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 *c1 = l1,*c2 = l2;
ListNode *d = new ListNode(0);
ListNode *headlsit = d;
int sum = 0;
while(c1 != NULL || c2 != NULL){
sum/=10;
if(c1 != NULL){
sum += c1->val;
c1 = c1->next;
}
if(c2 !=NULL){
sum += c2->val;
c2 = c2->next;
}
d->next = new ListNode(sum);
d = d->next;
}
if(sum/10==1)
d->next = new ListNode(1);
return headlsit->next;
}
};
