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。

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 }} /** * @param {ListNode} l1 * @param {ListNode} l2 * @return {ListNode} */const addTwoNumbers = (l1: ListNode, l2: ListNode) => { let l3: null | ListNode = null let cache: ListNode | null = null let tens: number = 0 while (l1 || l2) { let total: number = 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}

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 tens = 0 while l1 or l2: total = 0 if l1: l1Head = l1.val total = total + l1Head l1 = l1.next if l2: l1Head = l2.val total = total + l1Head l2 = l2.next total = total + tens if total >= 10: total = total - 10 tens = 1 else: tens = 0 node = ListNode(total) if cache: cache.next = node cache = node else: l3 = node cache = l3 if tens == 1: cache.next = ListNode(1) return l3

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