【Leetcode】Multiply Strings

最新推荐文章于 2025-08-08 21:26:37 发布

yyz20002008

最新推荐文章于 2025-08-08 21:26:37 发布

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文章标签： leetcode Java StringBuilder

本文介绍了如何在Java中实现两个大整数字符串的相乘操作，并提供了三种不同的实现方案，包括详细的代码示例和解释。

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iven two numbers represented as strings, return multiplication of the numbers as a string.

Note: The numbers can be arbitrarily large and are non-negative.

Java:

1. http://blog.youkuaiyun.com/fightforyourdream/article/details/17370495

这个好理解点

public class Solution {
    public String multiply(String num1, String num2) {
       // reverse
        String n1= new StringBuilder(num1).reverse().toString();
        
        String n2= new StringBuilder(num2).reverse().toString();
        
        
        int[] d= new int[n1.length()+n2.length()];//build array save product
        for(int i=0;i<n1.length();i++)
        {
            for(int j=0;j<n2.length();j++)
            {
                d[i+j]+=(n1.charAt(i)-'0')*(n2.charAt(j)-'0');//accumulate at correct position
            }
        }
        StringBuilder res = new StringBuilder();
        for(int i=0;i<d.length;i++)
        {
            int digit=d[i]%10;//current position
            int carry=d[i]/10;//carry position
            if(i+1<d.length)
            {
                d[i+1]+=carry;
            }
            res.insert(0,digit);
        }
        //remove first 0
        while(res.charAt(0)=='0'&&res.length()>1)
        {
            res.deleteCharAt(0);
        }
        
        return res.toString();
    }
}

2. Code大神这个for循环里面不是太好理解

public class Solution {
    public String multiply(String num1, String num2) {
        if(num1==null||num2==null||num1.length()==0||num2.length()==0) return null;
        if(num1.charAt(0)=='0') return "0";
        if(num2.charAt(0)=='0') return "0";
        StringBuilder res=new StringBuilder();
        int num=0;
        for(int i=num1.length()+num2.length();i>0;i--)
        {
            for(int j=Math.min(i-1,num1.length());j>0;j--)
            {
                if(i-j<=num2.length())
                {
                    num+=(int)(num1.charAt(j-1)-'0')*(int)(num2.charAt(i-1-j)-'0');
                }
                
            }
            if(i!=1||num>0) res.append(num%10);
            num=num/10;
        }
        return res.reverse().toString();
    }
}

3. 水印人生的数组法：http://gongxuns.blogspot.com/2013/01/leetcode-multiply-strings.html

public class Solution {
    public String multiply(String num1, String num2) {
        if(num1==null||num2==null||num1.length()==0||num2.length()==0) return "";
        if(num1.charAt(0)=='0') return "0";
        if(num1.charAt(0)=='0') return "0";
        int[] num=new int[num1.length()+num2.length()];
        for(int i=0;i<num1.length();i++){
            int carry=0;
            int a=num1.charAt(num1.length()-1-i)-'0';
            for(int j=0;j<num2.length();j++){
                int b=num2.charAt(num2.length()-1-j)-'0';
                num[i+j]+=carry+a*b;
                carry=num[i+j]/10;
                num[i+j]%=10;
            }
            num[i+num2.length()]+=carry;
        }
        int i=num.length-1;
        while(i>0&&num[i]==0) i--;
        StringBuilder res= new StringBuilder("");
        while(i>=0){
            res.append((char)('0'+num[i]));
            i--;
            
        }
        return res.toString();
    }
}