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leetCode解题记录2 - 两数相加(JS, TS, PY3版)

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 = None

class 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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