2. Add Two Numbers

1. 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.

2. 结果

2.1 golang版

type ListNode struct {
      Val int
      Next *ListNode
  }

func addTwoNumbers(l1 *ListNode, l2 *ListNode) *ListNode {
    temp := &ListNode{}
    result := temp
    p := l1
    q := l2
    carry := 0
    for p != nil && q != nil {
        temp.Next = &ListNode{}
        temp.Next.Val = (carry + p.Val + q.Val) %10
        carry = (carry + p.Val + q.Val) / 10
        p = p.Next
        q = q.Next
        temp = temp.Next
    }
    for p != nil{
        temp.Next =&ListNode{}
        temp.Next.Val = (carry + p.Val ) %10
        carry = (carry + p.Val ) / 10
        p = p.Next
        temp = temp.Next
    }
    for q != nil{
        temp.Next =&ListNode{}
        temp.Next.Val = (carry + q.Val ) %10
        carry = (carry + q.Val ) / 10
        q = q.Next
        temp = temp.Next
    }
    if carry != 0{
        temp.Next = &ListNode{}
        temp.Next.Val = carry
    }
    return result.Next
}

3. 分析

本题目其实是两个链表合并的问题。

1. 为了容易操作，我们先定义了result链表来表示新的结果链表，同时设置头指针。
2. 进行链表相加的操作。在进行链表相加时，当一个链表为空时。不能用if直接进行链表后续的连接，

    if q != nil{
        temp.Next =&ListNode{}
        temp.Next.Val = (carry + q.Val ) %10
        temp.Next.Next= q.Next
    }

因为，有可能存在1+99这种，会导致一直有进位相加,需要用for循环。
3.当两个链表都已经遍历到空时，需要注意如果最后有进位要把进位给加上。