Q2.5 Add Two Numbers

Q: You have two numbers represented by a linked list, where each node contains a single digit. The digits are stored in reverse order, such that the 1’s digit is wat the head of the list. Write a function that adds the two numbers and returns the sum as a linked list. EXAMPLE Input: (3 -> 1 -> 5) + (5 -> 9 -> 2) Output: 8 -> 0 -> 8

A: 简单的逐位相加问题，。不要溜掉最后以为有进位的情况即可。

#include <iostream>
using namespace std;

struct ListNode {
    int val;
	ListNode *next;
 	ListNode(int x) : val(x), next(NULL) {}
 };
 
ListNode *init(int a[], int n) {
 	ListNode *head = NULL;
 	ListNode *p = NULL;
 	for (int i = 0; i < n; i++) {
 		ListNode *cur = new ListNode(a[i]);
 		if (i == 0) {
 			head = cur;
 			p = cur;
 		}
 		p->next = cur;
 		p = cur;
 	}
 	return head;
}

ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
        if (!l1) {
            return l2;
        } else if (!l2) {
            return l1;
        }
        ListNode dummy(-1);
        int carry = 0;
        ListNode *prev = &dummy;
        for (ListNode *p = l1, *q = l2; p != NULL || q != NULL; p = p?p->next:NULL, q = q?q->next:NULL, prev = prev->next) {
            const int a = p?p->val:0;
            const int b = q?q->val:0;
            int value = (a+b+carry)%10;
            carry = (a+b+carry)/10;
            prev->next = new ListNode(value);
        }
        if (carry == 1) {
            prev->next = new ListNode(carry);
        }
        return dummy.next;
}

void printList(ListNode *head) {
	ListNode *p = head;
	for( ; p; p = p->next) {
		cout<<p->val<<" ";
	}
	cout<<endl;
}

int main() {
	int a[10] = {2,4,3};
	int b[10] = {5,6,4};
	ListNode *head1 = init(a, 3);
	ListNode *head2 = init(b, 3);
	printList(head1);
	printList(head2);
	ListNode *head = addTwoNumbers(head1, head2);
	printList(head);
	return 0;
}