# [LeetCode] 2. Add Two Numbers 两个数字相加

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.

Example:

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

C++ 解法：

class Solution {
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
ListNode *dummy = new ListNode(-1), *cur = dummy;
int carry = 0;
while (l1 || l2) {
int val1 = l1 ? l1->val : 0;
int val2 = l2 ? l2->val : 0;
int sum = val1 + val2 + carry;
carry = sum / 10;
cur->next = new ListNode(sum % 10);
cur = cur->next;
if (l1) l1 = l1->next;
if (l2) l2 = l2->next;
}
if (carry) cur->next = new ListNode(1);
return dummy->next;
}
};

Java 解法：

public class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
ListNode dummy = new ListNode(-1);
ListNode cur = dummy;
int carry = 0;
while (l1 != null || l2 != null) {
int d1 = l1 == null ? 0 : l1.val;
int d2 = l2 == null ? 0 : l2.val;
int sum = d1 + d2 + carry;
carry = sum >= 10 ? 1 : 0;
cur.next = new ListNode(sum % 10);
cur = cur.next;
if (l1 != null) l1 = l1.next;
if (l2 != null) l2 = l2.next;
}
if (carry == 1) cur.next = new ListNode(1);
return dummy.next;
}
}

Github 同步地址：

https://github.com/grandyang/leetcode/issues/2

Multiply Strings