Add two numbers represented by linked list in Go (Golang)

Table of Contents

Overview

Assume that both the linked list is already reversed. Meaning that let’s say the numbers are

The output should be 817

Assume both numbers are represented as a linked list in a reverse manner

8->7->4
9->3->3

And expected output linked list is also reversed

7->1->8

Here is the program for the same

package main

import "fmt"

func main() {
	first := initList()	
        first.AddFront(4)
	first.AddFront(7)
	first.AddFront(8)

	second := initList()
	second.AddFront(3)
	second.AddFront(3)
	second.AddFront(9)

	result := addTwoNumbers(first.Head, second.Head)
	result.Traverse()
}

func initList() *SingleList {
	return &SingleList{}
}

type ListNode struct {
	Val  int
	Next *ListNode
}

func (l *ListNode) Traverse() {
	for l != nil {
		fmt.Println(l.Val)
		l = l.Next
	}
}

type SingleList struct {
	Len  int
	Head *ListNode
}

func (s *SingleList) AddFront(num int) {
	ele := &ListNode{
		Val: num,
	}
	if s.Head == nil {
		s.Head = ele
	} else {
		ele.Next = s.Head
		s.Head = ele
	}
	s.Len++
}

func addTwoNumbers(l1 *ListNode, l2 *ListNode) *ListNode {
	var prev *ListNode
	var result *ListNode
	carry := 0

	for l1 != nil && l2 != nil {

		sum := l1.Val + l2.Val + carry

		carry = sum / 10
		if sum >= 10 {
			sum = sum % 10
		}

		tempNode := &ListNode{Val: sum}

		if prev == nil {
			result = tempNode
			prev = tempNode
		} else {
			prev.Next = tempNode
			prev = prev.Next
		}

		l1 = l1.Next

		l2 = l2.Next
	}

	for l1 != nil {
		sum := l1.Val + carry
		carry = sum / 10
		if sum >= 10 {
			sum = sum % 10
		}
		tempNode := &ListNode{Val: sum}
		prev.Next = tempNode
		l1 = l1.Next
	}

	for l2 != nil {
		sum := l2.Val + carry
		carry = sum / 10
		if sum >= 10 {
			sum = sum % 10
		}
		tempNode := &ListNode{Val: sum}
		prev.Next = tempNode
		l2 = l2.Next
	}

	if carry > 0 {
		tempNode := &ListNode{Val: carry}
		prev.Next = tempNode
	}
	return result
}

Output

7
1
8

golang