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Python练习:解题 - 两数相加(JS, TS, PY3版)Definition for singly-linked list.# class ListNode:# def __init__(self, x):# self.val = x# self.next = Noneclass Solution:

两数相加

题目描述

给出两个 非空 的链表用来表示两个非负的整数。其中,它们各自的位数是按照 逆序 的方式存储的,并且它们的每个节点只能存储 一位 数字。

如果,我们将这两个数相加起来,则会返回一个新的链表来表示它们的和。

您可以假设除了数字 0 之外,这两个数都不会以 0 开头。

输入:(2 -> 4 -> 3) + (5 -> 6 -> 4)

输出:7 -> 0 -> 8

原因:342 + 465 = 807

解题思路

其实这题比较简单,无非是两个链表之间同层级的数字相加,唯一要注意的就是如果相加之后数字大于10,需要往下一级+1,当前级数是个位的那个数字。基本也是一个循环可以解决的。再注意处理下,如果一个链表长度长于另一个链表时的边界处理,其余就没啥了。

Python练习:解题 - 两数相加(JS, TS, PY3版)Definition for singly-linked list.# class ListNode:# def __init__(self, x):# self.val = x# self.next = Noneclass Solution:

Python练习:解题 - 两数相加(JS, TS, PY3版)

JS版

/**

  • @param {ListNode} l1
  • @param {ListNode} l2
  • @return {ListNode}

    */const addTwoNumbers = (l1, l2) => {

let l3 = null

let cache = 0

let tens = 0

while (l1 || l2) {

let total = 0

if (l1) {

let l1Head = l1.val

total += l1Head

l1 = l1.next }

if (l2) {

let l2Head = l2.val

total += l2Head

l2 = l2.next }

total += tens if (total >= 10) {

total -= 10

tens = 1

} else {

tens = 0

}

let node = new ListNode(total)

if (cache) {

cache.next = node

cache = node } else {

l3 = node

cache = l3 }

if (tens === 1) {

cache.next = new ListNode(1)

return l3}

TS版

class ListNode {

val: number

next: ListNode | any

constructor(value: number) {

this.val = value this.next = null

}}

/**

  • */const addTwoNumbers = (l1: ListNode, l2: ListNode) => {

let l3: null | ListNode = null

let cache: ListNode | null = null

let tens: number = 0

let total: number = 0

PY版

Definition for singly-linked list.# class ListNode:# def __init__(self, x):# self.val = x# self.next = Noneclass Solution:

def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:

"""

:type l1: ListNode

:type l2: ListNode

:rtype: ListNode

l3 = None

cache = 0

while l1 or l2:

total = 0

if l1:

l1Head = l1.val

total = total + l1Head

l1 = l1.next

if l2:

l1Head = l2.val

l2 = l2.next

total = total + tens if total >= 10:

total = total - 10

else:

node = ListNode(total)

if cache:

cache = node else:

cache = l3 if tens == 1:

cache.next = ListNode(1)

return l3