leetcode 43. Multiply Strings

1. Multiply Strings Add to List QuestionEditorial Solution My Submissions
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   Given two numbers represented as strings, return multiplication of the numbers as a string.

Note:
The numbers can be arbitrarily large and are non-negative.
Converting the input string to integer is NOT allowed.
You should NOT use internal library such as BigInteger.
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这道题花了3个小时。
主要是一开始完全没有思路。以为可以转成自己把string转integer来做，后来发现数值大的离谱。
想来想去，还是得用小学的乘法。
乘法是按照N*M这种循环。
难点1：在于carry（进数）的处理。
最里层的循环乘完后，必须要处理其carry。否则carry会丢失。

这道题还是挺考察基本的循环编程能力的，可惜我状态不是很好。

柳儿说心疼我站阳台给她数绵羊 -_-, 满满的幸福。

/**
 * @param {string} num1
 * @param {string} num2
 * @return {string}
 */
var multiply = function(num1, num2) {


    if (num1.length < 1 || num2.length < 1) {return new string(0);}
    var sign = 1;
    if (num1[0]=='-'&& num2[0]=='+' || num1[0]=='+'&& num2[0]=='-') { sign = -1}

    var result = new Array(num1.length + num2.length);
    result.fill(0);
    for(var i = num2.length-1 ; i>= 0 && num2[i] <= '9' && num2[i] >= '0' ; i--)
    {   
        var carry = 0;
        for(var j = num1.length - 1; j >= 0 && num1[j] <= '9' && num1[j] >= '0' ; j--)
        {
            sum = (num2[i] - '0') * (num1[j] - '0') + result[i+j+1] +carry;
             result[i+j+1] = sum % 10 ;
             carry = Math.floor(sum /10);
            // console.log("sum:"+sum+", i+j+1:"+(i+j+1)+" result["+(i+j+1)+"]:"+result[i+j+1])

        }
        result[i+j+1] = carry;//卡在这步了，愣是分析不出来
    }
    //
  //  console.log("result:"+result);
    var resultString = "";

    var k = 0; 
    while (result[k] === 0 ) { k++;}
    for(; k< result.length ; k++)
    {
        resultString += result[k];
    }
    if (resultString === "") return "0"
    return resultString;
};