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";
    }