Add Two Numbers

You are given two linked lists representing two non-negative numbers. 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.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

　　遍历两个链表，对每个节点依次相加即可，最后遍历完两个链表之后还要注意进位的值是否为0.

　　c++实现：

ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
int carry = 0;
ListNode *head = NULL,*p = NULL;
while(l1||l2){
if(l1){
carry += l1->val;
ListNode *temp = l1;
l1 = l1->next;
free(temp);
}
if(l2){
carry += l2->val;
ListNode *temp = l2;
l2 = l2->next;
free(temp);
}
ListNode *temp = new ListNode(carry%10);
carry /= 10;
if(p){
p->next = temp;
p = p->next;
}
else{
head = p = temp;
}
}
if(carry){
p->next = new ListNode(carry);
}
return head;
}

　　java实现：

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
ListNode ret = null;
ListNode p = null;
int carry = 0;
while(l1!=null||l2!=null){
int x = 0,y = 0,sum = 0;
if(l1!=null){
x = l1.val;
l1 = l1.next;
}
if(l2!=null){
y = l2.val;
l2 = l2.next;
}
sum = x+y+carry;
carry = sum/10;
ListNode temp = new ListNode(sum%10);
if(ret==null){
ret = temp;
p = ret;
}
else{
p.next = temp;
p = p.next;
}
}
if(carry>0){
p.next = new ListNode(carry);
}
return ret;
}