167. 链表求和

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判断两个链表的大小，如果链表长度不想等：以短的链表为准，遍历链表计算。然后再遍历长链表。

count来记录是否进位，如果两个链表都遍历完，count还有进位，那么再建立一个节点存1，来表示多出来的进位。

class Solution {
public:
/*
* @param l1: the first list
* @param l2: the second list
* @return: the sum list of l1 and l2
*/
ListNode * addLists(ListNode * l1, ListNode * l2) {
// write your code here
int count1 = 0, count2 = 0;
ListNode *p1 = l1;
ListNode *p2 = l2;
ListNode *head;
while(p1 != NULL)
{
count1++;
p1 = p1->next;
}
while(p2 != NULL)
{
count2++;
p2 = p2->next;
}
if(count1 < count2)
{
head = add(l1, l2);
}
else
{
head = add(l2, l1);
}
return head;
}
ListNode *add(ListNode *l1, ListNode *l2)
{
if(l1 == NULL && l2 == NULL) return l1;
ListNode *head = NULL;
ListNode *p, *q;
int num = 1;
int count = 0;
while(l1 != NULL)
{
num = l1->val + l2->val + count;
count = 0;
if(num >= 10)
{
num = num % 10;
count = 1;
}
p = new ListNode(num);
if(head == NULL)
{
head = p;
q = head;
}
else
{
q->next = p;
q = q->next;
}
l1 = l1->next;
l2 = l2->next;
}
while(l2 != NULL)
{
num = l2->val + count;
count = 0;
if(num >= 10)
{
num = num % 10;
count = 1;
}
p = new ListNode(num);
if(head == NULL)
{
head = p;
q = head;
}
else
{
q->next = p;
q = q->next;
}
l2 = l2->next;
}
if(count == 1)
{
num += count;
if(num >= 10)
{
num = num % 10;
count = 1;
}
p = new ListNode(num);
if(head == NULL)
{
head = p;
q = head;
}
else
{
q->next = p;
q = q->next;
}
}
return head;
}
};

我还是认为我的代码不够简洁，希望还可以找到更好的方法。

2018/1/26