golang 算法题 : 两数相加

最新推荐文章于 2024-11-18 15:51:01 发布

转载最新推荐文章于 2024-11-18 15:51:01 发布 · 224 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/w3liu/p/10850325.html

文章标签：

#golang

本文介绍了一种使用Go语言实现的链表相加算法，通过定义链表节点结构和addTwoNumbers函数，实现了两个链表数值的相加。算法处理了进位问题，并能正确输出相加后的链表数值。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

package main

import "fmt"

type ListNode struct {
   Val  int
 Next *ListNode
}

func main() {
   l1 := &ListNode{
      Val: 1,
 Next: &ListNode{
         Val: 2,
 Next: &ListNode{
            Val: 3,
 },
 },
 }
   l2 := &ListNode{
      Val: 4,
 Next: &ListNode{
         Val: 5,
 Next: &ListNode{
            Val: 6,
 },
 },
 }
   l3 := addTwoNumbers(l1, l2)
   num := ""
 for l3 != nil {
      val := l3.Val
      num = fmt.Sprintf("%s%d", num, val)
      l3 = l3.Next
   }
   fmt.Println(num)
}

func addTwoNumbers(l1 *ListNode, l2 *ListNode) *ListNode {
var head *ListNode
 var previous *ListNode
 var current *ListNode
 var carry int
 for l1 != nil || l2 != nil {
      v1 := 0
 v2 := 0
 if l1 != nil {
         v1 = l1.Val
      }
if l2 != nil {
         v2 = l2.Val
      }
      result := v1 + v2 + carry
      nodeVal := result % 10
 carry = result / 10
 current = &ListNode{
         Val: nodeVal,
 }
if head == nil {
         head = current
      }
if previous != nil {
         previous.Next = current
      }

      previous = current

if l1 != nil {
         l1 = l1.Next
      }
if l2 != nil {
         l2 = l2.Next
      }

   }

if carry > 0 {
      current = &ListNode{
         Val: carry,
 }
if head == nil {
         head = current
      }
if previous != nil {
         previous.Next = current
      }
   }

return head
}