leetcode:Multiply Strings

最新推荐文章于 2019-01-11 00:33:46 发布

原创最新推荐文章于 2019-01-11 00:33:46 发布 · 327 阅读

0 ·

CC 4.0 BY-SA版权

算法同时被 3 个专栏收录

537 篇文章

订阅专栏

C++/C

504 篇文章

订阅专栏

leetcode

230 篇文章

订阅专栏

本文详细阐述了如何使用字符串表示的任意大小的非负数进行乘法运算，并通过算法实现将其转换为字符串形式的结果。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

class Solution {
public:
    
    void reverse(string &str)
    {
        if (str.size() == 0)
            return;
            
        int beginIdx = 0;
        int endIdx   = str.size()-1;
        
        while (beginIdx<endIdx)
        {
            int tmp = str.at(beginIdx);
            str[beginIdx] = str[endIdx];
            str[endIdx] = tmp;
            
            beginIdx++;
            endIdx--;
        }
    }

    string multiply(string num1, string num2) {
        
        reverse(num1);
        reverse(num2);
        vector<int> result(num1.size()+num2.size());
        int lastPos;
        
        for (int i=0; i<num1.size(); i++)
        {
            for (int j=0; j<num2.size(); j++)
            {
                int tmp = (num1[i]-'0') * (num2[j]-'0');
                result[i+j] += tmp;
            }
        }
        
        int carry = 0;
        string retValue="";
        for (int i=0; i<result.size(); i++)
        {
            int tmp = result[i]+carry;
        
            if (tmp >= 10)
            {
                result[i] = tmp%10;
                carry = tmp/10;
            }
            else
            {
                result[i] = tmp;
                carry = 0;
            }
        
            retValue = (char)(result[i]+'0') + retValue;
        }
        
        int i=0;
        for (; i<retValue.size(); i++)
        {
            if (retValue[i] != '0')
            {
                retValue = retValue.substr(i, retValue.size()-i);
                break;
            }
        }
        
        if (i == retValue.size() && retValue[retValue.size()-1] == '0')
            return "0";
        else
            return retValue;
    }

};