题目
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
即：342 + 465 = 807
我的思路
此题主要要求熟悉链表（可以参考我的一篇译文），理解了链表的定义、遍历和添加节点等方法之后，很容易可以得到结果。注意需要判断当输入的两个链表有可能长度不一致的情况。
代码
struct ListNode
* addTwoNumbers(struct ListNode
* l1, struct ListNode
* l2) {
int tmp
= 0;
struct ListNode
* retls
= NULL;
retls
= (struct ListNode
*)malloc(sizeof(struct ListNode));
if(retls
== NULL) {
printf(
"malloc failed!\n");
return retls;
}
retls
->val
= 0;
retls
->next
= NULL;
struct ListNode
*curl1
= l1;
struct ListNode
*curl2
= l2;
struct ListNode
*curret
= retls;
while((curl1
!= NULL)
&& (curl2
!= NULL)) {
tmp
= curret
->val
+ curl1
->val
+ curl2
->val;
curret
->val
= tmp
% 10;
if((curl1
->next
== NULL)
&& (curl2
->next
== NULL)
&& (tmp
< 10)) {
curret
->next
= NULL;
}
else {
curret
->next
= (struct ListNode
*)malloc(sizeof(struct ListNode));
curret
->next
->val
= tmp
/ 10;
curret
->next
->next
= NULL;
curret
= curret
->next;
}
curl1
= curl1
->next;
curl2
= curl2
->next;
}
while(curl1
!= NULL) {
tmp
= curret
->val
+ curl1
->val;
curret
->val
= tmp
% 10;
if((curl1
->next
== NULL)
&& (tmp
< 10)) {
curret
->next
= NULL;
}
else {
curret
->next
= (struct ListNode
*)malloc(sizeof(struct ListNode));
curret
->next
->val
= tmp
/ 10;
curret
->next
->next
= NULL;
curret
= curret
->next;
}
curl1
= curl1
->next;
}
while(curl2
!= NULL) {
tmp
= curret
->val
+ curl2
->val;
curret
->val
= tmp
% 10;
if((curl2
->next
== NULL)
&& (tmp
< 10)) {
curret
->next
= NULL;
}
else {
curret
->next
= (struct ListNode
*)malloc(sizeof(struct ListNode));
curret
->next
->val
= tmp
/ 10;
curret
->next
->next
= NULL;
curret
= curret
->next;
}
curl2
= curl2
->next;
}
return retls;
}
