题目
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.
给定两个非空链表来表示两个非负整数。位数按照逆序方式存储，它们的每个节点只存储单个数字。将两数相加返回一个新的链表。
你可以假设除了数字 0 之外，这两个数字都不会以零开头。
示例：
输入：(2 -> 4 -> 3) + (5 -> 6 -> 4)
输出：7 -> 0 -> 8
原因：342 + 465 = 807
解题思路
先新建一个链表的虚拟头节点，然后同时开始遍历两个已知的链表，拿出两个数计算之和，注意点是需要记录一下上一次之后有无进位；然后把结果一次放到新建的链表后面，遍历完成之后，把新链第一个虚拟头结点去掉返回链表即可
当然，大家也可以暴力一点，直接遍历两个已知链表，把结果取出来，倒序相加之后，在倒序放入链表返回；不过官方解法似乎使用的是上一种
解法
# ruby解法
# Definition for singly-linked list.
# class ListNode
# attr_accessor :val, :next
# def initialize(val)
# @val = val
# @next = nil
# end
# end
# @param {ListNode} l1
# @param {ListNode} l2
# @return {ListNode}
def add_two_numbers(l1, l2)
dummy_head = ListNode.new(0)
curr = dummy_head
carry = 0
while (l1 != nil || l2 != nil) do
num1 = l1 ? l1.val : 0
num2 = l2 ? l2.val : 0
l1 = l1.next if l1
l2 = l2.next if l2
sum = num1 + num2 + carry
curr.next = ListNode.new(sum % 10)
carry = sum >= 10 ? 1 : 0
curr = curr.next
end
curr.next = ListNode.new(1) if carry > 0
dummy_head.next
end
/**
* java解法
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
ListNode dummyHead = new ListNode(0);
ListNode curr = dummyHead;
int carry = 0;
while(l1 != null || l2 != null) {
int num1 = l1 != null ? l1.val : 0;
int num2 = l2 != null ? l2.val : 0;
int sum = num1 + num2 + carry;
carry = sum / 10;
curr.next = new ListNode(sum % 10);
curr = curr.next;
if (l1 != null) l1 = l1.next;
if (l2 != null) l2 = l2.next;
}
if (carry > 0) {
curr.next = new ListNode(carry);
}
return dummyHead.next;
}