求助:调用链表函数报错,求完整调用流程

两个数字相加完整代码调用

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

链表相加函数:

package main

import (
    "container/list"
    "fmt"
)

type ListNode struct {
    Val int
    Next *ListNode
}

func main() {
    //创建一个list
    l1 := list.New()
    l1.PushBack(2)
    l1.PushBack(4)
    l1.PushBack(3)

    l2 := list.New()
    l2.PushBack(4)
    l2.PushBack(3)

    for e := l1.Front(); e != nil; e = e.Next() {
        fmt.Println(e.Value)
    }
    for e := l2.Front(); e != nil; e = e.Next() {
        fmt.Println(e.Value)
    }

    //调用addTwoNumbers函数

}

func addTwoNumbers(l1 *ListNode, l2 *ListNode) *ListNode {
    head := &ListNode{0, nil}
    current := head
    carry := 0
    for l1 != nil || l2 != nil || carry > 0 {
        sum := carry
        if l1 != nil {
            sum += l1.Val
            l1 = l1.Next
        }
        if l2 != nil {
            sum += l2.Val
            l2 = l2.Next
        }
        carry = sum / 10
        current.Next = new(ListNode)
        current.Next.Val = sum % 10
        current = current.Next
    }
    return head.Next
}
讨论数量: 2

可以关注我的公众号 golang那点事,里面得算法专栏有这道题得实现源码,以及多种解题思路

4年前 评论
awesee
/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func addTwoNumbers(l1 *ListNode, l2 *ListNode) *ListNode {
    r := &ListNode{}
    t := r
    carry := 0
    for l1 != nil || l2 != nil || carry != 0 {
        val := carry
        if l1 != nil {
            val += l1.Val
            l1 = l1.Next
        }
        if l2 != nil {
            val += l2.Val
            l2 = l2.Next
        }
        t.Next = &ListNode{Val: val % 10}
        t = t.Next
        carry = val / 10
    }
    return r.Next
}

源码链接

4年前 评论

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