leetCode解題記錄2 - 兩數相加（JS TS PY版）

時間 2019-11-05

標籤 leetcode 解題記錄相加欄目應用數學简体版

原文原文鏈接

做者：陳大魚頭

項目地址：ying-leetcode

碎碎念：Mmmmm，不按期刷leetcode，會以JS TS PY的形式輸出出來

題目描述

給出兩個非空的鏈表用來表示兩個非負的整數。其中，它們各自的位數是按照逆序的方式存儲的，而且它們的每一個節點只能存儲一位數字。javascript

若是，咱們將這兩個數相加起來，則會返回一個新的鏈表來表示它們的和。java

您能夠假設除了數字 0 以外，這兩個數都不會以 0 開頭。node

輸入：(2 -> 4 -> 3) + (5 -> 6 -> 4) 輸出：7 -> 0 -> 8 緣由：342 + 465 = 807python

解題思路

其實這題比較簡單，無非是兩個鏈表之間同層級的數字相加，惟一要注意的就是若是相加以後數字大於10，須要往下一級+1，當前級數是個位的那個數字。基本也是一個循環能夠解決的。再注意處理下，若是一個鏈表長度長於另外一個鏈表時的邊界處理，其他就沒啥了。git

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
}
複製代碼

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
}
複製代碼

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
複製代碼