字符串相乘

给定两个以字符串形式表示的非负整数 num1 和 num2，返回 num1 和 num2 的乘积，它们的乘积也表示为字符串形式。

注意：不能使用任何内置的 BigInteger 库或直接将输入转换为整数。

示例 1:

输入: num1 = "2", num2 = "3"
输出: "6"

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/**
 * @param {string} num1
 * @param {string} num2
 * @return {string}
 */
var multiply = function(num1, num2) {
    if(num1 === "0" || num2 === "0") return "0"
    let len1 = num1.length
    let len2 = num2.length
    let maxLen = 0

    let arr = []
    for(let i = len1 - 1; i >=0; i--) {
        let curArr = new Array(len1 - i - 1).fill(0)
        let intPart = 0
        for(let j = len2 - 1; j >=0; j--) {
            let curRes = num1[i] * num2[j]

            if(j === 0 && curRes + intPart >= 10) {
                curArr.unshift(Math.floor((curRes + intPart)/10),(curRes + intPart)%10)
                intPart = 0
            } else {
                curArr.unshift((curRes + intPart)%10)
                intPart = Math.floor((curRes + intPart)/10)
            }
        }

        arr.push(curArr)
    }


    maxLen = arr[arr.length - 1].length

    for(let m = 0; m < arr.length; m++) {
        if(arr[m].length < maxLen) {
            while(arr[m].length < maxLen) {
                arr[m].unshift(0)
            }
        }
    }

    let res = []
    let resIntPart = 0

    for(let n = maxLen - 1; n >= 0; n--) {
        let curRes = 0
        for(let m = 0; m < arr.length; m++) {
            curRes += arr[m][n]
        }
        if(n === 0 && curRes + resIntPart >= 10) {
            res.unshift(Math.floor((curRes + resIntPart)/10),(curRes + resIntPart)%10)
        } else {
            res.unshift((curRes + resIntPart)%10)
            resIntPart = Math.floor((curRes + resIntPart)/10)
        }
    }

    
    return res.join("")

};

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来源：力扣（LeetCode）链接：https://leetcode-cn.com/problems/multiply-strings

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