[LeetCode] 059: Multiply Strings

[Problem]

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.

[Solution]

class Solution {
public:
/**
 * calculate the sum of the numbers
 */
string add(string num1, string num2){
string res = "";
int sum = 0, carry = 0;
int i = num1.size() - 1, j = num2.size() - 1;
while(i >= 0 || j >= 0){
sum = carry;
if(i >= 0){
sum += num1[i--] - '0';
}
if(j >= 0){
sum += num2[j--] - '0';
}
carry = sum / 10;
sum %= 10;
res = string(1, char(sum + '0')) + res;
}
if(carry > 0){
res = string(1, char(carry + '0')) + res;
}

// remove head '0'
i = 0;
while(i < res.size() && res[i] == '0')i++;

// return
if(i == res.size()){
return "0";
}
else{
return res.substr(i, res.size() - i);
}
}
/**
 * multiply the two strings
 */
string multiply(string num1, string num2) {
// Note: The Solution object is instantiated only once and is reused by each test case.
vector<string> numbers;
string tail = "";
for(int i = num2.size()-1; i >= 0; --i){
string tmp = "";
int carry = 0, res = 0;
for(int j = num1.size()-1; j >= 0; --j){
res = carry + (num2[i] - '0') * (num1[j] - '0');
carry = res / 10;
res %= 10;
tmp = string(1, char(res + '0')) + tmp;
}
if(carry > 0){
tmp = string(1, char(carry + '0')) + tmp;
}
tmp += tail;
tail += "0";
numbers.push_back(tmp);
}

// add the numbers
string sum = "0";
for(int i = 0; i < numbers.size(); ++i){
sum = add(sum, numbers[i]);
}
return sum;
}
};

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