Java for LeetCode 043 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.

解题思路一：

BigInteger！！！ JAVA实现如下：

import java.math.BigInteger;

public class Solution {
static public String multiply(String num1, String num2) {
BigInteger bi1=BigInteger.valueOf(0);
for(int i=0;i<num1.length();i++){
bi1=bi1.multiply(BigInteger.valueOf(10));
bi1=bi1.add(BigInteger.valueOf(num1.charAt(i)-'0'));
}
BigInteger bi2=BigInteger.valueOf(0);
for(int i=0;i<num2.length();i++){
bi2=bi2.multiply(BigInteger.valueOf(10));
bi2=bi2.add(BigInteger.valueOf(num2.charAt(i)-'0'));
}
return bi1.multiply(bi2).toString();
}
}

360 ms Accepted，值得注意的是系统不会自动导入math包，需要在声明时添加。

解题思路二：使用BigInteger总有种作弊的赶脚，题目的真正意思是让我们用加法模拟乘法的过程，JAVA实现如下：

static public String multiply(String num1, String num2) {
StringBuilder result=new StringBuilder();
for(int i=0;i<num1.length()+num2.length();i++)
result.append('0');
for (int i = num1.length()-1; i >=0; i--)
for (int j = num2.length()-1; j>=0; j--) {
int tmp = (result.charAt(i+j+1) - '0') + (num1.charAt(i) - '0') * (num2.charAt(j) - '0');
result.replace(i+j+1, i+j+2, ""+tmp % 10);
for(int k=i+j;;k--){
tmp=result.charAt(k)- '0' + tmp / 10;
result.replace(k, k+1,""+ tmp%10);
if(tmp<10)
break;
}
}
for (int i = 0; i <result.length(); i++)
if(result.charAt(i)!='0')
return result.substring(i, result.length());
return "0";
}