[LeetCode-Java]43. Multiply Strings

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.

解：为了避免溢出，显然应该考虑对应位相乘，求结果的每一位的值最后串在一起。
比如，217*561，个位=7*1，十位=1*1+7*6，百位 = 7*5+1*6+2*1，千位=2*6+1*5，万位=2*5。分别存进数组里面，在考虑低位向高位的进位。转换为字符串之后考虑首位为0的情况。

public class Solution {    public String multiply(String num1, String num2) {        //字符串反转        num1 = new StringBuilder(num1).reverse().toString();        num2 = new StringBuilder(num2).reverse().toString();        //两数相乘  最大位数为两者位数之和        int[] b = new int[num1.length()+num2.length()];        for (int i = 0;i < num1.length();i++){            for (int j = 0;j < num2.length();j++){                //先获取对应位相乘  不考虑进位                b[i+j] += (num1.charAt(i) - '0') * (num2.charAt(j) - '0');            }        }        for (int i = 0;i<num1.length() + num2.length();i++){            //对进位进行处理            if (b[i]>9){                b[i+1] += b[i]/10;                b[i] %= 10;            }        }        StringBuilder stringBuilder = new StringBuilder();        for (int i = num1.length()+num2.length()-1;i>=0;i--){            stringBuilder.append(b[i]);        }        //去除首部的0        while (stringBuilder.charAt(0) == '0' && stringBuilder.length()>1)  stringBuilder.deleteCharAt(0);        return stringBuilder.toString();    }}