LeeCode_MultiplyStrings

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经过昨天一天，还有今天的学习，终于是把第一个程序给搞出来，虽然是借鉴他人的，但是还是收获颇多。现在把

这个程序的问题，和解决方案在下面贴出：

MultiplyStrings

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.

给定用字符串表示的两个数字，计算其乘积：

public class MultiplyStrings {//这种算法是模拟了正常的乘积运算解出答案。

<span style="white-space:pre"></span>//原始解法中问号部分，我认为是不需要的，不太理解是什么意思，

<span style="white-space:pre"></span>//我在Leecode中测试是可以通过的。public static void main(String[] args) { String a1 = "12888"; String b1 = "13131555";String  s = multiply(a1, b1);System.out.println("s =" + s); } public static String multiply(String num1, String num2) { StringBuilder s1 = new StringBuilder(num1).reverse();    StringBuilder s2 = new StringBuilder(num2).reverse();    int l1 = s1.length();    int l2 = s2.length();    int[] result = new int[l1 + l2 - 1];    for(int i = 0; i < l1; i++){        for(int j = 0; j < l2; j++){            result[i+j] += (s1.charAt(i) - '0') * (s2.charAt(j) - '0');        }    }    //System.out.println(Arrays.toString(result));    StringBuilder s = new StringBuilder();    int carry = 0;    for(int i = 0; i < result.length; i++){        int currDigit = (result[i] + carry) % 10;        carry = (result[i]+carry)/10;        s.insert(0, currDigit);        System.out.println(s);    }    if(carry!=0){        s.insert(0, carry);    }/*???????????????????????????????????????????????????????    int i = 0;    while( i < s.length() && s.charAt(i) == '0'){        i++;    }    if(i ==  s.length() ) s.delete(0,i-1);    else s.delete(0, i);*/    return s.toString(); }}

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