<LeetCode OJ> 43. Multiply Strings

43. Multiply Strings

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以字符串的形式给定两个数字，返回相乘的结果，注意：结果也是字符串，因为数字可能很大

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.

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分析：

代码注释已经很详细，模拟手算即可

//思路首先:
//模拟手算过程即可
class Solution {
public:
    string multiply(string num1, string num2) {
        int allLen=num1.size()+num2.size();
        vector<int> tmpresult(allLen,0);
        string result(allLen,'0');
        //模拟手算从最后一位开始处理
        for(int i=num1.size()-1;i>=0;i--)
        {
            int n1=num1[i]-'0';
            for(int j=num2.size()-1;j>=0;j--)
            {
                int n2=num2[j]-'0';
                tmpresult[i+j+1]+= n1*n2;
            }
        }
        //进位
        for(int i=allLen-1;i>0;i--)
        {
            while(tmpresult[i]>9)
            {
                tmpresult[i-1]+=tmpresult[i]/10;
                tmpresult[i]%=10;
            }
        }
        //转换成字符串
        for(int i=allLen-1;i>=0;i--)
            result[i]=tmpresult[i]+'0';

        if(result.find_first_not_of('0') == string::npos)
            return "0";//排除全0的情况
        return result.substr(result.find_first_not_of('0'),string::npos);
    }
};

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原文地址：http://blog.csdn.net/ebowtang/article/details/50575524

原作者博客：http://blog.csdn.net/ebowtang