[Leetcode]2. Add Two Numbers

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

大意为：给定两个链表，链表中的值为非负整数，将两个链表中的值相加

思路：

从list1, list2头节点开始，对应位置相加并建立新节点。用一个变量carry记录进位。注意几种特殊情况：

1. 一个链表为空
   l1: NULL
   l2: 1->2
   sum: 1->2

2. l1, l2长度不同，且结果有可能长度超过l1, l2中的最大长度
   l1: 2->2
   l2: 9->9->9
   sum: 1->2->0->1

// in swift
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public var val: Int
 *     public var next: ListNode?
 *     public init(_ val: Int) {
 *         self.val = val
 *         self.next = nil
 *     }
 * }
 */
class Solution {
    func addTwoNumbers(_ l1: ListNode?, _ l2: ListNode?) -> ListNode? {
        var carry = 0
        var list1 = l1, list2 = l2
        let dummy = ListNode(0)
        var pre  = dummy
        
        while list1 != nil || list2 != nil {
            let val1: Int = list1?.val ?? 0
            let val2: Int = list2?.val ?? 0
            
            let val = val1 + val2 + carry
            
            pre.next = ListNode(val % 10)
            pre = pre.next!
            carry = val >= 10 ? 1: 0
            list1 = list1?.next
            list2 = list2?.next
        }
        
        if carry != 0 {
            pre.next = ListNode(1)
        }
        
        return dummy.next
    }
}