LeetCode | 43. Multiply Strings(大整数乘法)

Given two non-negative integers num1 and num2 represented as strings, return the product of num1 and num2.

Note:The length of both num1 and num2 is < 110.Both num1 and num2 contains only digits 0-9.Both num1 and num2 does not contain any leading zero.You must not use any built-in BigInteger library or convert the inputs to integer directly.

大整数乘法，6ms AC

#include <iostream>#include <string>using namespace std;string multiply(string num1, string num2){    //i 和 j 相乘的结果是 res 的 (i+j) 位    string str_res = "";    int first = 0;    int res[300] = {};    int len1 = num1.length(), len2 = num2.length();    if(len1 < len2)        return multiply(num2, num1);    int a[120] = {};    int b[120] = {};    for(int i=0;i<len1;i++)        a[i] = num1[len1-i-1]-'0';    for(int i=0;i<len2;i++)        b[i] = num2[len2-i-1]-'0';    for(int i=0;i<len1;i++)    {        for(int j=0;j<len2;j++)        {            res[i+j] += a[i]*b[j];        }    }    for(int i=0;i<len1+len2+2;i++)    {        if(res[i] >= 10)        {            res[i+1] += res[i]/10;            res[i] %= 10;        }    }    for(int i=290;i>=0;i--)        if(res[i] != 0)        {            first = i; break;        }    for(int i=first;i>=0;i--)    {        str_res += res[i]+'0';    }    return str_res;}int main(){    string str1, str2;    while(cin>>str1>>str2)    {        cout<<multiply(str1, str2)<<endl;    }    return 0;}